RD Sharma solution class 7 chapter 15 Properties of triangles Exercise 15.5

Exercise 15.5

Page-15.30

Question 1:

State Pythagoras theorem and its converse.

Answer 1:

The Pythagoras Theorem: In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.

Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.

Question 2:

In right ∆ ABC, the lengths of the legs are given. Find the length of the hypotenuse.
(i) a = 6 cm, b = 8 cm
(ii) a = 8 cm, b = 15 cm
(iii) a = 3 cm, b = 4 cm
(iv) a = 2 cm, b = 1.5 cm

Answer 2:

According to the Pythagoras theorem,
$(\mathrm{Hypotenuse}{)}^2 = (\mathrm{Base}{)}^2 + (\mathrm{Height}{)}^2$

$$\begin{gathered} \left(\mathrm{i}\right) {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \\ {\mathrm{c}}^2=6^2+8^2 \\ {\mathrm{c}}^2=36+64=100 \\ \mathrm{c}=10 \mathrm{cm} \\ \left(\mathrm{ii}\right) {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \\ {\mathrm{c}}^2=8^2+{15}^2 \\ {\mathrm{c}}^2=64+225=289 \\ \mathrm{c}=17 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \\ {\mathrm{c}}^2=3^2+4^2 \\ {\mathrm{c}}^2=9+16=25 \\ \mathrm{c}=5 \mathrm{cm} \\ \left(\mathrm{iv}\right) {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \\ {\mathrm{c}}^2=2^2+1.5^2 \\ {\mathrm{c}}^2=4+2.25=6.25 \\ \mathrm{c}=2.5 \mathrm{cm} \end{gathered}$$

Question 3:

The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm, find the length of the other side.

Answer 3:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{hypotenuse} \mathrm{be} "\mathrm{c}" \mathrm{and} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{sides} \mathrm{be} "\mathrm{b}" \mathrm{and} "\mathrm{c}". \\ \mathrm{Using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \\ 2.5^2=1.5^2+{\mathrm{b}}^2 \\ {\mathrm{b}}^2=6.25-2.25=4 \\ \mathrm{b}=2 \\ \mathrm{Hence}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{side} \mathrm{is} 2 \mathrm{cm}. \end{gathered}$$

Question 4:

A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.

Answer 4:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{hypotenuse} \mathrm{be} \mathrm{h}. \\ \mathrm{Using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{get}: \\ 3.7^2=1.2^2+{\mathrm{h}}^2 \\ {\mathrm{h}}^2=13.69-1.44=12.25 \\ \mathrm{h}=3.5 \\ \mathrm{Hence}, \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{wall} \mathrm{is} 3.5 \mathrm{m}. \end{gathered}$$

Question 5:

If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle.

Answer 5:

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{triangle}, \mathrm{the} \mathrm{largest} \mathrm{side} \mathrm{is} 6 \mathrm{cm}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{in} \mathrm{a} \mathrm{right}-\mathrm{angled} \mathrm{triangle}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{squares} \mathrm{of} \mathrm{the} \mathrm{smaller} \mathrm{sides} \mathrm{should} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{square} \mathrm{of} \mathrm{the} \mathrm{largest} \mathrm{side}. \\ \mathrm{Therefore}, \\ {3}^2+4^2=9+16=25 \\ \mathrm{But}, \\ {6}^2=36 \\ \Rightarrow 3^2+4^2\ne 6^2 \\ \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{right}-\mathrm{angled} \mathrm{triangle}. \end{gathered}$$

Question 6:

The sides of certain triangles are given below. Determine which of them are right triangles.
(i) a = 7 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm

Answer 6:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{in} \mathrm{a} \mathrm{right}-\mathrm{angled} \mathrm{triangle}, \mathrm{the} \mathrm{square} \mathrm{of} \mathrm{the} \mathrm{largest} \mathrm{side} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{squares} \mathrm{of} \mathrm{the} \mathrm{smaller} \mathrm{sides}. \\ \mathrm{Here}, \mathrm{the} \mathrm{larger} \mathrm{side} \mathrm{is} \mathrm{c}, \mathrm{which} \mathrm{is} 25 \mathrm{cm}. \\ {\mathrm{c}}^2=625 \\ \mathrm{We} \mathrm{have}: \\ {\mathrm{a}}^2+{\mathrm{b}}^2=7^2+{24}^2=49+576=625={\mathrm{c}}^2 \\ \mathrm{Thus}, \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{triangle}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{in} \mathrm{a} \mathrm{right}-\mathrm{angled} \mathrm{triangle}, \mathrm{the} \mathrm{square} \mathrm{of} \mathrm{the} \mathrm{largest} \mathrm{side} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{squares} \mathrm{of} \mathrm{the} \mathrm{smaller} \mathrm{sides}. \\ \mathrm{Here}, \mathrm{the} \mathrm{larger} \mathrm{side} \mathrm{is} \mathrm{c}, \mathrm{which} \mathrm{is} 18 \mathrm{cm}. \\ {\mathrm{c}}^2=324 \\ \mathrm{We} \mathrm{have}: \\ {\mathrm{a}}^2+{\mathrm{b}}^2=9^2+{16}^2=81+256=337\ne {\mathrm{c}}^2 \\ \mathrm{Thus}, \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{right} \mathrm{triangle}. \end{gathered}$$

Question 7:

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
[Hint: Find the hypotenuse of a right triangle having the sides (11 − 6) m = 5 m and 12 m]

Answer 7:

$$\begin{gathered} \mathrm{The} \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{tops} \mathrm{of} \mathrm{the} \mathrm{poles} \mathrm{is} \mathrm{the} \mathrm{distance} \mathrm{between} \mathrm{points} \mathrm{A} \mathrm{and} \mathrm{B}. \\ \mathrm{We} \mathrm{can} \mathrm{see} \mathrm{from} \mathrm{the} \mathrm{given} \mathrm{figure} \mathrm{that} \mathrm{points} \mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{form} \mathrm{a} \mathrm{right} \mathrm{triangle}, \mathrm{with} \mathrm{AB} \mathrm{as} \mathrm{the} \mathrm{hypotenuse}. \\ \mathrm{On} \mathrm{using} \mathrm{the} \mathrm{Pythagoras} \mathrm{Theorem} \mathrm{in} △\mathrm{ABC}, \mathrm{we} \mathrm{get}: \\ (11-6{)}^2+{12}^2={\mathrm{AB}}^2 \\ \Rightarrow 25+144={\mathrm{AB}}^2 \\ \Rightarrow {\mathrm{AB}}^2=169 \\ \Rightarrow \mathrm{AB}=13 \\ \mathrm{Hence}, \mathrm{the} \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{tops} \mathrm{of} \mathrm{the} \mathrm{poles} \mathrm{is} 13 \mathrm{m}. \end{gathered}$$

Question 8:

A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

Answer 8:

$$\begin{gathered} \mathrm{Let} \mathrm{O} \mathrm{be} \mathrm{the} \mathrm{starting} \mathrm{point} \mathrm{and} \mathrm{P} \mathrm{be} \mathrm{the} \mathrm{final} \mathrm{point}. \\ \mathrm{By} \mathrm{using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{can} \mathrm{find} \mathrm{the} \mathrm{distance} \mathrm{OP}. \\ {15}^2+8^2={\mathrm{OP}}^2 \\ \Rightarrow {\mathrm{OP}}^2=225+64 \\ \Rightarrow {\mathrm{OP}}^2=289 \\ \Rightarrow \mathrm{OP}=17 \\ \mathrm{Hence}, \mathrm{the} \mathrm{required} \mathrm{distance} \mathrm{is} 17 \mathrm{m}. \end{gathered}$$

Page-15.31

Question 9:

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?

Answer 9:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{ladder} \mathrm{be} \mathrm{L} \mathrm{m}. \\ \mathrm{By} \mathrm{using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{can} \mathrm{find} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{ladder}. \\ {6}^2+8^2={\mathrm{L}}^2 \\ \Rightarrow {\mathrm{L}}^2=36+64=100 \\ \Rightarrow \mathrm{L}=10 \\ \mathrm{Thus}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{ladder} \mathrm{is} 10 \mathrm{m}. \end{gathered}$$


$$\begin{gathered} \mathbf{When}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{ladder}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{shifted}\mathbf{:} \\ \mathrm{Let} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{ladder} \mathrm{after} \mathrm{it} \mathrm{is} \mathrm{shifted} \mathrm{be} \mathrm{H} \mathrm{m}. \\ \mathrm{By} \mathrm{using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{can} \mathrm{find} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{ladder} \mathrm{after} \mathrm{it} \mathrm{is} \mathrm{shifted}. \\ {8}^2+{\mathrm{H}}^2={10}^2 \\ \Rightarrow {\mathrm{H}}^2=100-64=36 \\ \Rightarrow \mathrm{H}=6 \\ \mathrm{Thus}, \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{ladder} \mathrm{is} 6 \mathrm{m}. \end{gathered}$$

Question 10:

A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm. How far is the lower end of the ladder from the base of the wall?

Answer 10:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{lower} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{ladder} \mathrm{from} \mathrm{the} \mathrm{wall} \mathrm{be} \mathrm{X} \mathrm{m}. \\ \mathrm{On} \mathrm{using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{get}: \\ {\mathrm{X}}^2+{48}^2={50}^2 \\ \Rightarrow {\mathrm{X}}^2={50}^2-{48}^2=2500-2304=196 \\ \Rightarrow \mathrm{X}=14 \mathrm{dm} \\ \mathrm{Hence}, \mathrm{the} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{lower} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{ladder} \mathrm{from} \mathrm{the} \mathrm{wall} \mathrm{is} 14 \mathrm{dm}. \end{gathered}$$

Question 11:

The two legs of a right triangle are equal and the square of the hypotenuse is 50. Find the length of each leg.

Answer 11:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{each} \mathrm{leg} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{triangle} \mathrm{be} \mathrm{x} \mathrm{units}. \\ \mathrm{Using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{get}: \\ {\mathrm{x}}^2+{\mathrm{x}}^2=(\mathrm{Hypotenuse}{)}^2 \\ {\mathrm{x}}^2+{\mathrm{x}}^2=50 \\ 2{\mathrm{x}}^2=50 \\ \Rightarrow {\mathrm{x}}^2=25 \\ \Rightarrow \mathrm{x}=5 \\ \mathrm{Hence}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{each} \mathrm{leg} \mathrm{is} 5 \mathrm{units}. \end{gathered}$$

Question 12:

Verify that the following numbers represent Pythagorean triplet:
(i) 12, 35, 37
(ii) 7, 24, 25
(iii) 27, 36, 45
(iv) 15, 36, 39

Answer 12:

We will check for a Pythagorean triplet by checking if the square of the largest side is equal to the sum of the squares of the other two sides.

$$\begin{gathered} \left(\mathrm{i}\right) {37}^2=1369 \\ {12}^2+{35}^2=144+1225=1369 \\ {12}^2+{35}^2={37}^2 \\ \mathrm{Yes}, \mathrm{they} \mathrm{represent} \mathrm{a} \mathrm{Pythagorean} \mathrm{triplet}. \\ \left(\mathrm{ii}\right) {25}^2=625 \\ {7}^2+{24}^2=49+576=625 \\ {7}^2+{24}^2={25}^2 \\ \mathrm{Yes}, \mathrm{they} \mathrm{represent} \mathrm{a} \mathrm{Pythagorean} \mathrm{triplet}. \\ \left(\mathrm{iii}\right) {45}^2=2025 \\ {27}^2+{36}^2=729+1296=2025 \\ {27}^2+{36}^2={45}^2 \\ \mathrm{Yes}, \mathrm{t}\mathrm{hey} \mathrm{represent} \mathrm{a} \mathrm{Pythagorean} \mathrm{triplet}. \\ \left(\mathrm{iv}\right) {39}^2=1521 \\ {15}^2+{36}^2=225+1296=1521 \\ {15}^2+{36}^2={39}^{2 } \\ \mathrm{Yes}, \mathrm{they} \mathrm{represent} \mathrm{a} \mathrm{Pythagorean} \mathrm{t}\mathrm{riplet}. \end{gathered}$$

Question 13:

In a ∆ABC, ∠ABC = 100°, ∠BAC = 35° and BD ⊥ AC meets side AC in D. If BD = 2 cm, find ∠C and length DC.

Answer 13:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°. \\ \mathrm{Therefore}, \mathrm{for} \mathrm{the} \mathrm{given} △\mathrm{ABC}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180° \\ 100°+35°+\angle \mathrm{ACB}=180° \\ \Rightarrow \angle \mathrm{ACB}=180°-135° \\ \Rightarrow \angle \mathrm{ACB}=45° \\ \angle \mathrm{C}=45° \\ \mathrm{If} \mathrm{we} \mathrm{apply} \mathrm{the} \mathrm{above} \mathrm{rule} \mathrm{on} △\mathrm{BCD}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that}: \\ \angle \mathrm{BCD}+\angle \mathrm{BDC}+\angle \mathrm{CBD}=180° \\ 45°+90°+\angle \mathrm{CBD}=180° (\because \angle \mathrm{ACB}=\angle \mathrm{BCD} \mathrm{and} \mathrm{BD}\perp \mathrm{AC}) \\ \Rightarrow \angle \mathrm{CBD}=180°-135° \\ \Rightarrow \angle \mathrm{CBD}=45° \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles} \mathrm{have} \mathrm{equal} \mathrm{length}. \\ \mathrm{Thus}, \\ \mathrm{BD}=\mathrm{DC} \\ \mathrm{DC}=2 \mathrm{cm} \end{gathered}$$

Question 14:

In a ∆ABC, AD is the altitude from A such that AD = 12 cm, BD = 9 cm and DC = 16 cm. Examine if ∆ABC is right angled at A.

Answer 14:

$$\begin{gathered} \mathrm{In} △\mathrm{ADC}, \\ \angle \mathrm{ADC}=90° (\mathrm{AD} \mathrm{is} \mathrm{an} \mathrm{altitude} \mathrm{on} \mathrm{BC}.) \\ \mathrm{Using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{get}: \\ {12}^2+{16}^2={\mathrm{AC}}^2 \\ {\mathrm{AC}}^2=144+256=400 \\ \mathrm{AC}=20 \mathrm{cm} \\ \mathrm{In} △\mathrm{ADB}, \\ \angle \mathrm{ADB}=90° (\mathrm{AD} \mathrm{is} \mathrm{an} \mathrm{altitude} \mathrm{on} \mathrm{BC}.) \\ \mathrm{Using} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{get}: \\ {12}^2+9^2={\mathrm{AB}}^2 \\ {\mathrm{AB}}^2=144+81=225 \\ \mathrm{AB}=15 \mathrm{cm} \\ \mathrm{In} △\mathrm{ABC}, \\ {\mathrm{BC}}^2={25}^2=625 \\ {\mathrm{AB}}^2+{\mathrm{AC}}^2={15}^2+{20}^2=625 \\ {\mathrm{AB}}^2+{\mathrm{AC}}^2={\mathrm{BC}}^2 \\ \mathrm{Because} \mathrm{it} \mathrm{satisfies} \mathrm{the} \mathrm{Pythagoras} \mathrm{theorem}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} △\mathrm{ABC} \mathrm{is} \mathrm{right} \mathrm{angled} \mathrm{at} \mathrm{A}. \end{gathered}$$

Question 15:

Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 105°. Measure AB. Is (AB)² = (AC)² + (BC)²? If not, which one of the following is true:
(AB)² > (AC)² + (BC)² or (AB)² < (AC)² + (BC)²?

Answer 15:

Draw $△ABC$.
Draw a line BC = 3 cm.
At point C, draw a line at 105$°$ angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB, which will be approximately 5.5 cm.
$$\begin{gathered} A{C}^2+B{C}^2=4^2+3^2=9+16=25 \\ A{B}^2=5.5^2=30.25 \end{gathered}$$

(AB)² $\ne$ (AC)² + (BC)²

Here,

(AB)² > (AC)² + (BC)²

Question 16:

Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 80°. Measure AB. Is (AB)² = (AC)² + (BC)²? If not, which one of the following is true:
(AB)² > (AC)² + (BC)² or (AB)² < (AC)² + (BC)²?

Answer 16:

First draw $△ABC$.

Draw a line BC = 3 cm.
At point C, draw a line at 80$°$ angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB; it will be approximately 4.5 cm.
$$\begin{gathered} A{C}^2+B{C}^2=4^2+3^2=9+16=25 \\ A{B}^2=4.5^2=20.25 \end{gathered}$$

(AB)² $\ne$ (AC)² + (BC)²

Here,

(AB)² < (AC)² + (BC)²