SChand CLASS 9 Chapter 10 Rectilinear Figures Exercise 10(A)

 Exercise 10 A

(1) In a right angled triangle $A B C, c$ is the length of the hypotenuse, and $a$ and $b$ are other two sides.

(i) If $\mathrm{a}=6$ and $\mathrm{b}=8$, then find $\mathrm{c}$.

(ii) If $c=25$ and $a=24$, then find $b$.

(iii) If $c=13$ and $b=5$, them find $a$.

(iv) If $\mathrm{a}=10$ and $\mathrm{c}=21$, then find $\mathrm{b}$.

(v) if $a=9$ and $b=9$, then find $c$.

(2) A rectangular field is $30 \mathrm{~m}$ by $40 \mathrm{~m}$. What distance is saved by walking diagonally across the field?

(3) A man travels $7 \mathrm{~km}$ due north, then goes $3 \mathrm{~km}$ due east and then $3 \mathrm{~km}$ due south. how far is he form his starting point?

(4) The diagonals of a rhombus are $12 \mathrm{~cm}$ and $9 \mathrm{~cm}$ long. Calculate the length of one side of the rhombus.

(5) (i) In a right-angled triangle $A B C$ it is given that the hypotenuse, $A C=2.5 \mathrm{~cm}$, and the side $A B$ $=1.5 \mathrm{~cm} .$ Calculate the side BC.

(ii) $A D$ is drawn perpendicular to $B C$, base of an equilateral triangle $A B C$. Given $B C=10 \mathrm{~cm}$, find the length of $A D$, correct to one decimal place of decimal.

(6) $A B C$ is right angled triangle. Angle $A B C=90$ degree, $A C=25 \mathrm{~cm}$ and $A B=24 \mathrm{~cm}$. Calculate the area of $\triangle A B C$.

(7) A ladder $13 \mathrm{~m}$ long tests against a vertical wall, If the foot of the ladder is $5 \mathrm{~m}$ from the foot of the wall, find the distance of the other end of the ladder from ground.

(8) Use the given information to make a neat diagram of the figure having given properties and write down the name of the figures.

In triangle $A B C, A B^{2}=B C^{2}+A C^{2}$ and $A C=2 B C$.

(9) In fig $10.10, \mathrm{ABCD}$ represents a quadrilateral in which $\mathrm{AD}=13 \mathrm{~cm}, \mathrm{DC}=12 \mathrm{~cm}, \mathrm{BC}=3 \mathrm{~cm}$, and $\angle A B D=\angle B C D=90^{\circ} .$ Calculate the length of $A B .$

(10) Which of the triangles whose sides are given are given below are right angled?

(i) $4 \mathrm{~cm}, 5 \mathrm{~cm}, 6 \mathrm{~cm}$

(ii) $1.2 \mathrm{~cm}, 3.7 \mathrm{~cm}, 3.5 \mathrm{~cm}$.

(iii) $4 \mathrm{~cm}, 9.6 \mathrm{~cm}, 10.4 \mathrm{~cm}$

(iv) $2.2 \mathrm{~cm}, 3.3 \mathrm{~cm}, 4.4 \mathrm{~cm}$.

(11) In Fig. 10.11, find the distance of D from $A$, unit of length is $\mathrm{cm},(A B=2, B C=4, C D=2)$

(12) The sides of a right angled triangle containing the right angle are $5 x$ and $(3 x-1) \mathrm{cm}$. If the area of the triangle be $60 \mathrm{~cm}$ square, calculate the lengths of the sides of the triangle.

(13) In Fig. $10.12$ its is given that $A B=B C=25 \mathrm{~cm}, A E=7 \mathrm{~m}$, and $C D=24 \mathrm{~m}$, find the length of $D E$ and also show that triangle $A B E$ and triangle $B D C$ are congruent.

(14) A ladder rests against a vertical wall at a height of $12 \mathrm{~m}$ from the ground with its foot at a distance of $9 \mathrm{~m}$ from the wall on the ground. If the foot of the ladder is shifted $3 \mathrm{~m}$ away from the wall, how much lower will the ladder slide down?

(15) $A D$ is an altitude of a $\triangle A B C$ and $A D$ is $12 \mathrm{~cm} B D=9 \mathrm{~cm}$ and $D C=16 \mathrm{~cm}$ long respectively. Prove that the angle BAC is a right angle.

(16) The shortest distance AP from a point A to a straight line QR is $12 \mathrm{~cm}$, and Q, R are $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$ distance from $A$ on opposite sides of $A P$, prove that $Q A R$ is a right angle.

(17) In Fig. 10.13, PT is an altitude of the triangle PQR, In which PQ $-25 \mathrm{~cm}, \mathrm{PR}=17 \mathrm{~cm}, \mathrm{PT}=15$ $\mathrm{cm}$ and $\mathrm{QR}=\mathrm{x} \mathrm{cm}$. Calculate $\mathrm{x}$.

(18) In Fig. 10.14, $A B=8 \mathrm{~cm}, B C=6 \mathrm{~cm}, A C=3 \mathrm{~cm}$ and the angle $A D C=90$ degree, calculate $C D$.

(19) In Fig. 10.15, the angle $B A C$ is a right angle and $A D$ is perpendicular to $B C ; A B=4 \mathrm{~cm}$ and $A C$ $=3 \mathrm{~cm}$ and $\mathrm{BD}=\mathrm{x}$, Calculate $\mathrm{x}$

(20) In fig. 10.16, the angle BAD and ADC are right angles and AX is parallel to BC. If AB $=B C=5$ $\mathrm{cm}$, and $\mathrm{DC}=8 \mathrm{~cm}$, Calculate the area of $\mathrm{ABCX} .$

SOLUTIONS 1

(i)

$\begin{aligned} c &=\sqrt{a^{2}+b^{2}} \\ &=\sqrt{6^{2}+8^{2}} \\ &=\sqrt{36+64} \\ &=\sqrt{100}=10 \end{aligned}$

(ii)

$\begin{aligned} b &=\sqrt{(25)^{2}-(24)^{2}} \\ &=\sqrt{625-576} \\ &=\sqrt{49}=\sqrt{7} \end{aligned}$

(iii)

$\begin{aligned} a &=\sqrt{(13)^{2}-5^{2}} \\ &=\sqrt{169-25} \\ &=\sqrt{144} \\ &=12 \end{aligned}$

(iv)$b=\sqrt{(29)^{2}-(16)^{2}}$

$=\sqrt{441-100}$

$=\sqrt{ 341}$

(v)

$\begin{aligned} C=& \sqrt{9^{2}\left(9\right)^{2}} \\ &=\sqrt{182} \\ &=9 \sqrt{2} m \end{aligned}$

 Solution 2:

 Distance Saved = $\sqrt{(30)^{2}+(40)^{2}} \mathrm{~m}$

$\begin{aligned} &=\sqrt{900+1600} \\ &=\sqrt{2500} \mathrm{~m} \\ &=50 \mathrm{~m} \end{aligned}$

Solution 3:  (IMAGE TO BE ADDED)

Distance from starting point= $\sqrt{4^{2}+3^{2}}$

$=\sqrt{49+9}$ 

=$\sqrt{16+9}$

=5m

 Solution 4

(IMAGE TO BE ADDED)

$A B=\sqrt{(9 / 2)^{2}+6^{2}} \mathrm{~cm}^{2}$

$=\sqrt{\frac{81}{4}+36}\mathrm{cm}^{2}$

$=\sqrt{\frac{144}{4}+81}$

$\sqrt{\frac{225}{4}}$

$=\frac{15}{2}=$7.5 

Solution 5

 (i) (IMAGE TO BE ADDED)

BC = $\sqrt{( \sqrt{25})^{2}-(1 \cdot 5)^{2}} \mathrm{~cm}$

$\sqrt{6.25}-2.25$

$=\sqrt{4} \mathrm{~cm}$

$=2 \mathrm{~cm}$

(ii)(IMAGE TO BE ADDED)

$A D=\sqrt{(10)^{2}-5^{2}} \mathrm{~cm}$

$=\sqrt{100-25} \mathrm{~cm}$

$=\sqrt{75} \mathrm{~cm}$

$=8.7 \mathrm{~cm}$

Solution 6

(IMAGE TO BE ADDED)

$B C=\sqrt{(25)^{2}-(29)^{2}}$

$=\sqrt{625-576 \mathrm{~cm}}$

=$\sqrt{49 \mathrm{~cm}}$

=7cm

Arua $=\frac{1}{2} \times 7 \times 24 \mathrm{~cm}^{2}$

$=84 \mathrm{~cm}^{2}$

Solution 7

(IMAGE TO BE ADDED)

$A B=\sqrt{(13)^{2}-5^{2}} m$

=$\sqrt{169-25} \mathrm{~m}$

=$\sqrt{144}$

=12 m

Solution 8

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$A B^{2}=A C^{2}+B C^{2}$

$A C=2 B C$

Name of the figure in right angle triangle with right angle at C

Solution 9

(IMAGE TO BE ADDED)

$\begin{aligned} D B^{2} &=(12)^{2} +(3)^{2} \\ &=144+2 \\ D B &=\sqrt{153} \mathrm{~cm} \end{aligned}$

$\begin{aligned} A B=& \sqrt{169-153} \\=& 4 \mathrm{~cm} \end{aligned}$

Solution 10

(i) $6^{2}=5^{2}+4^{2}$

$36= 25+16$ Not right angle triangle 

(ii)$(3 \cdot 7)^{2}=(3 \cdot 5)^{2}+(\sqrt{2})^{2}$ 

So, it is a right angle triangle ,

(iii)$(10 \cdot 4)^{2}=(9 \cdot 6)^{2}+4^{2}$

right angle triangle 

(iv)$(4.4)^{2} \neq(3.3)^{2}+(2.2)^{2}$

Not a right angle triangle 

Solution 11

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Solution 12

$\frac{1}{2} \times 5 x \times(3 x-1)=60$

=$x(3 x-1)=24$

=$3 x^{2}-x-24=0$

=$3 x^{2}-9 x+8 x-24=0$

=$3 x(x-3)+8(x-3)=0$

$(x-3)(2 x+8)=0$ 

Length of the side is positive so sides are 

$5 \times 3=15 \mathrm{~cm}$ $=(3 \times 3-1)=8 \mathrm{~cm}$

=$\sqrt{(15)^{2}+64}$ 

=$\sqrt{225+64}$

$\sqrt{289}=$ 17cm

Solution 13

(IMAGE TO BE ADDED)

$\begin{aligned} D B &=\sqrt{(25)^{2}-(24)^{2}} \mathrm{~m} \\ &=\sqrt{625-576} \mathrm{~m} \\ &=\sqrt{49} \mathrm{~m} \\ &=7 \mathrm{~m} \end{aligned}$

$\begin{aligned} B C &=\sqrt{(25)^{2}-7^{2}} \\ &=\sqrt{625-49} \\ &=\sqrt{576} \mathrm{~m} \\ &=24 \mathrm{~m} \end{aligned}$

So, ADBC & ABCD 

Length of DE= (24+7)= 31m

Solution 14

(IMAGE TO BE ADDED)

$A C^{2}=(12)^{2}+9^{2}=144+81$ = 225 

$A C=15 \mathrm{~m}$

$B E=\sqrt{(15)^{2}-12^{2}}$ =$\sqrt{225-144}$

$=\sqrt{81} \mathrm{~m}$

=9m

Ladder side down = (12-9)=3m

Solution 15

(IMAGE TO BE ADDED)

$\triangle A B D$ is right ang le triangle, 

$\begin{aligned} A B &=\sqrt{12^{2}+9^{2}} \mathrm{~cm} \\ &=\sqrt{144+81} \mathrm{~cm} \\ &=\sqrt{225} \mathrm{~cm} \\ &=15 \mathrm{~cm}\end{aligned}$

$\begin{aligned} A C &=\sqrt{144+256} \\ &=\sqrt{400} \mathrm{~cm} \\ &=20 \mathrm{~cm} \end{aligned}$

$B C=(3+16)=25 \mathrm{~cm}$

$\therefore(25)^{2}=(20)^{2}+(15)^{2}$

$\therefore \triangle B A C$ is right angle triangle 

Solution 16

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(not given)

Solution 17

(IMAGE TO BE ADDED)

$\begin{aligned} Q T &=\sqrt{(25)^{2}-(15)^{2}} \mathrm{~cm} \\ &=\sqrt{(325-22 \pi} \mathrm{cm} \\ &=\sqrt{400} \mathrm{~cm} \\ &=20 \mathrm{~cm} \end{aligned}$

$\begin{aligned} R T &=\sqrt{(17)^{2}-(15)^{2}} \mathrm{~cm} \\ &=\sqrt{289-225} \mathrm{~cm} \\ &=\sqrt{64} \mathrm{~cm} \\ &=8 \mathrm{~cm} \end{aligned}$

$\because x=Q R=(20+8)=28 \mathrm{~cm}$

Solution 18

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$C D=x \mathrm{~cm}$

$\triangle A B D$ We get, 

$A B^{2}=B D^{2}+A D^{2}$

${S}^{2}=(6+n)^{2}+A D^{2}$

From triangle ,  $\triangle A B D$ ................(i)

$3^{2}=AD^{2}+x^{2}$...........................(ii)

From (i) & (ii) We get 

$8^{2}-(6+x)^{2}=3^{2}-x^{2}$

$64=\left(36+12 x+x^{2}\right)$=$3^{2}-x^{2}$

$28-12 x-x^{2}$=$9-x^{2}$

$12 x=19$

$x=17 / 12 \mathrm{~cm}$

Solution 19

(IMAGE TO BE ADDED)

$B C=\sqrt{4^{2}+3^{3}

$=\sqrt{16+2}=5 \mathrm{~cm}$

$B C=\sqrt{4^{2}+3^{2}}=$$\sqrt{25}=5 \mathrm{~cm}$

DC = 5-x

$A D=\sqrt{16-x^{2}}(From$\triangle AB D$)$

$A D=\sqrt{9-(5-x)^{2}}$ (From  $\triangle AD C$ )

$\left(6-x^{2}=9-(5-x)^{2}\right.$

$7=x^{2}-$ (25-2 x 5.X + $x^{2}$)

$7=x^{2}-25+10 x$-$x^{2}$

$x=3 \cdot 2$

Solution 20

(IMAGE TO BE ADDED)

From $\triangle A D X$ We get , 

$\begin{aligned} 5^{2}=& 3^{2}+A D^{2} \\ A D^{2} &=25-9 \\ &=16 \end{aligned}$

AD = 4cm

Area of $\triangle B C E$= $\frac{1}{2} \times 3 \times 4$

= $6 \mathrm{~cm}^{2}$

Area of ABCX = Area of ABED+ Area of BCE - Area of ADX

= $5 \times 4+$ 6 -6

$=20 \mathrm{~cm}^{2}$