RS Aggarwal 2019,2020 solution class 7 chapter 17 Construction Exercise Test Paper 17

Test Paper 17

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Question 1:

In the given figure, AB || CD, ∠ABO = 60° and ∠CDO = 40°. Then, find ∠BOD.

Answer 1:

$\begin{array}{l}\\ \angle \text{ABO}={60}^{{}^0}\\ \angle {\text{CDO =40}}^0\\ \angle \text{ABO}=\angle \text{BOC }\left[\text{alternate angles}\right]\\ {\text{ =60}}^0\\ \angle \text{CDO }=\angle \text{DOC}={40}^0\left[\text{alternate angles}\right]\\ \angle \text{BOD}=\angle \text{BOC}+\angle \text{DOC}={60}^0+{40}^0={100}^0\\ \end{array}$

Question 2:

In the given figure, CE || BA. If ∠BAC = 70° and ∠ECD = 50°, find ∠ACB.

Answer 2:

$\begin{array}{l}\text{Here, AB II EC}\\ \therefore \angle \text{BAC}=\angle \text{ACE}={70}^0 (\mathrm{alternate} \mathrm{angles})\\ \angle \text{BCA+}\angle \text{ACE+}\angle \text{ECD=18}{0}^0 \\ \angle \text{BCA=18}{0}^0-{120}^0\\ \angle \text{BCA}={60}^0\end{array}$

Question 3:

In the given figure, two straight lines AB and CD intersect at a point O such that ∠AOC = 50°.
Find: (i) ∠BOD (ii) ∠BOC.

Answer 3:

$\begin{array}{l}\\ \left(\text{i}\right)\angle \text{AOC =}\angle {\text{BOD =50}}^0\text{ [vertically opposite angles]}\\ \left(\text{ii}\right)\angle \text{BOC }={180}^0-{50}^0 (\mathrm{linear} \mathrm{pair})\\ {\text{ =130}}^0\\ \end{array}$

Question 4:

In the given figure, AOB is a straight line and OC is ray such that∠AOC = (3x + 20)° and ∠BOC = (2x − 10)°. Find the value of x and hence find (i) ∠AOC and ∠BOC.

Answer 4:

$\begin{array}{l}\text{Here, 3}x+20+2x-10=180\\ =>5x+10=180\\ \text{ =>5}x=170\\ =>x=34\\ \angle \text{AOC}=(3\times 34+20)°\\ \text{ =}\left(\text{102}+20\right)\text{°}\\ \text{ =122}°\\ \angle \text{BOC =(2}\times \text{34}-10)°\\ \text{ =(68}-10)°\\ =58°\end{array}$

Question 5:

In a ∆ABC, If ∠A = 65°, ∠B = 45°, find ∠C.

Figure

Answer 5:

$\begin{array}{l}\text{In }\Delta \text{ABC, }\angle \text{A+}\angle \text{B+}\angle {\text{C=180}}^0\\ {\text{ =>65}}^0+{45}^0+\angle \text{C}={180}^0\\ \text{ =>}\angle {\text{C=180}}^0-{110}^0={90}^0\end{array}$

Question 6:

In the given figure, x : y = 2 : 3 and ∠ACD = 120°. Find the values of x,y and z.

Answer 6:

$\begin{array}{l}\text{Let }x=2k\text{ and y=3}k\\ \therefore 2k+3k={120}^0 (e\mathrm{xterior} \mathrm{angle} \mathrm{property})\\ =>5k={120}^0\\ =>k={24}^0\\ \therefore x=2\times {24}^0={48}^0\text{ and y=3}\times {\text{24}}^0={72}^0\\ \text{In }\Delta \text{ABC:}\\ \angle \text{A}+\angle \text{B}+\angle \text{C}={180}^0\\ =>{48}^0+{72}^0+\angle {\text{C=180}}^0\\ \text{ =>}\angle \text{C}={180}^0-{120}^0\\ \text{ =>}\angle \text{C}={60}^0\\ \therefore z={60}^0\end{array}$

Question 7:

Two legs of a right triangle are 8 cm and 15 cm long. Find the length of the hypotenuse of the triangle.

Answer 7:

$$\begin{gathered} \text{Since it is a right triangle, by using the Pythagoras theorem:} \\ \begin{array}{l}\text{Length of the hypotenuse}=\sqrt{8^2+{15}^2}\\ =\sqrt{64+225}\\ =\sqrt{289}\\ =\pm 17 \text{cm}\\ \end{array} \\ \text{The length of the side can not be negative. } \\ \therefore \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{hypotenuse}=17 \mathrm{cm} \end{gathered}$$

Question 8:

In the adjoining figure, ABC is a triangle in which AD is the bisector of ∠A. If AD ⊥ BC, show that ∆ABC is isosceles.

Answer 8:

$$\begin{gathered} \begin{array}{l}\text{Given:}\\ \angle \text{BAD}=\angle \text{DAC ...(i)}\\ \text{ To show that }△\text{ABC is isoceles, we should show that }\angle \text{B}=\angle \text{C. }\\ \\ \therefore \text{AD}\perp \text{BC,}\angle \text{ADB}=\angle \text{ADC}=90{°}^{}\\ \end{array} \\ \angle ADC=\angle ADB \\ \angle BAD+\angle ABD=\angle DAC+\angle ACD (exterior angle property) \\ \angle DAC+\angle ABD=\angle DAC+\angle ACD (\mathrm{from} \mathrm{equation} (\mathrm{i}\left)\right) \\ \angle \mathrm{A}\mathrm{B}\mathrm{D}=\angle \mathrm{A}\mathrm{C}\mathrm{D} \\ \mathrm{This} \mathrm{is} \mathrm{because} \mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} ∆\mathrm{ABC} \mathrm{are} \mathrm{equal}. \\ \mathrm{Hence}, \mathrm{ABS} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \end{gathered}$$

Question 9:

Construct a ∆ABC in which BC = 5.3 cm, ∠B = 60° and AB = 4.2 cm. Also, draw the perpendicular bisector of AC.

Answer 9:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ \text{ 1}\text{. Draw BC}=5.3\text{ cm }\\ \text{ 2}\text{. Construct }\angle \text{CBX}=60{°}^{}\\ \text{ 3}\text{. With B as the centre and radius 4.2 cm, cut the ray BX at point A.}\\ \text{ 4}\text{. Join A and C}\text{.}\\ \text{ 5}\text{. With A as the centre and radius more than half of AC, draw an arc on either side of AC}\text{.}\end{array} \\ 6. \mathrm{With} \mathrm{C} \mathrm{as} \mathrm{the} \mathrm{centre} \mathrm{and} \mathrm{the} \mathrm{same} \mathrm{radius}, \mathrm{draw} \mathrm{another} \mathrm{arc} \mathrm{cutting} \mathrm{the} \mathrm{previously} \mathrm{drawn} \mathrm{arc} \mathrm{at} \mathrm{M} \mathrm{and} \mathrm{N}. \\ 7. \mathrm{Join} \mathrm{M} \mathrm{and} \mathrm{N}. \end{gathered}$$

Question 10:

Mark (✓) against the correct answer
The supplement of 35° is

(a) 55°
(b) 65°
(c) 145°
(d) 165°

Answer 10:

$$\begin{gathered} \left(\mathrm{c}\right) 145° \\ \text{Supplement of 35°= 180}°-35°=145° \end{gathered}$$

Question 11:

Mark (✓) against the correct answer
In the given figure, AOB is a straignt line, ∠AOC = 56° and ∠BOC = x°. The value of x is



(a) 34
(b) 44
(c) 144
(d) 124

Answer 11:

$$\begin{gathered} \left(d\right)124 \\ \begin{array}{l}\\ x^0+{56}^0={180}^0 (\mathrm{linear} \mathrm{pair})\\ \text{ => }{x}^0={180}^0-{56}^0\\ {\text{ =124}}^0\end{array} \\ \text{∴ x=124} \end{gathered}$$

Question 12:

Mark (✓) against the correct answer
In ∆ABC, side BC has been produced to D such that ∠ACD = 125° and ∠BAC = 60°. Then ∠ABC = ?



(a) 55°
(b) 60°
(c) 65°
(d) 70°

Answer 12:

$$\begin{gathered} \left(c\right) 65° \\ \begin{array}{l}\\ \angle \text{ACD}={125}^0\\ \angle ACD=\angle CAB+\angle ABC (\because \text{ the exterior angles are equal to the sum of its interior opposite angles})\\ \therefore \angle \text{ABC}={125}^0-{60}^0={65}^0\\ \end{array} \end{gathered}$$

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Question 13:

Mark (✓) against the correct answer
In a ∆ABC, If ∠B = 40° and ∠C = 35°, then ∠A = ?

(a) 50°
(b) 55°
(c) 105°
(d) 150°

Answer 13:

$$\begin{gathered} \left(c\right) 105° \\ \begin{array}{l}\angle \text{A+}\angle \text{B+}\angle {\text{C=180}}^0\\ \text{ =>}\angle \text{A}={\text{180}}^0-({40}^0+{35}^0)\\ \text{ =>}\angle \text{A}={\text{105}}^0\\ \end{array} \end{gathered}$$

Question 14:

Mark (✓) against the correct answer
In a ∆ABC, If 2∠A = 3∠B = 6∠C, then ∠B = ?

(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 14:

$$\begin{gathered} \begin{array}{l}\left(c\right) 60° \\ Given: \\ 2\angle A=3\angle B \\ \angle A=\frac{3}{2}\angle B ...\left(\mathrm{i}\right) \\ 3\angle B=3\angle C \\ \angle C=\frac{1}{2}\angle B ...\left(\mathrm{ii}\right)\\ \end{array} \\ \angle A+\angle B+\angle C=180° \\ \begin{array}{l}\text{ =>}\frac{3}{2}\angle \text{B}+\angle \text{B}+\frac{1}{2}\angle \text{B}={\text{180°}}^{}\\ \text{ =>}(\frac{3}{2}+1+\frac{1}{2})\angle \text{B}=\text{180°}\\ \text{ =>}\frac{6}{2}\angle \text{B}=180°\\ \text{ =>}\angle \text{B}=\frac{180°}{3}=60°\end{array} \end{gathered}$$

Question 15:

Mark (✓) against the correct answer
In a ∆ABC, If A − B = 33° and B − C = 18°, then ∠B = ?

(a) 35°
(b) 55°
(c) 45°
(d) 57°

Answer 15:

$$\begin{gathered} \left(\mathrm{b}\right) 55° \\ \begin{array}{l}\text{ In }△\text{ABC:}\\ \text{A+B+C}=180{°}^{} ... \left(\mathrm{i}\right)\\ Given, A-B=33°\end{array} \\ A=33°+B ...\left(\mathrm{ii}\right) \\ \mathrm{B}-\mathrm{C}=18° \\ \mathrm{C}=\mathrm{B}+18° ...\left(\mathrm{iii}\right) \\ \mathrm{Putting} \mathrm{the} \mathrm{value}s \mathrm{of} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{in} \mathrm{equation} \left(\mathrm{i}\right): \\ \Rightarrow \mathrm{B}+33°+\mathrm{B}+\mathrm{B}-18°=180° \\ \Rightarrow 3\mathrm{B}=180°-15 \\ \Rightarrow \mathrm{B}=\frac{165°}{3}=55° \end{gathered}$$

Question 16:

Mark (✓) against the correct answer
∆ABC is an isosceles right triangle in which ∠A = 90° and BC = 6 cm. Then AB = ?

(a) $2\sqrt{2}\mathrm{cm}$
(b) $3\sqrt{2}\mathrm{cm}$
(c) $4\sqrt{2}\mathrm{cm}$
(d) $2\sqrt{3}\mathrm{cm}$

Answer 16:

$$\begin{gathered} \left(b\right) 3\sqrt{2}\mathrm{cm} \\ \begin{array}{l}\text{ Here, AB}=\text{AC}\\ \text{ In right angled isoceles triangle: }\\ {\text{ BC}}^2={\text{AB}}^2+{\text{AC}}^2\\ {\text{ =>BC}}^2=2{\text{AB}}^2\\ \text{ =>}36=2{\text{AB}}^2\\ {\text{ =>AB}}^2=\frac{3{\sout{6}}^{18}}{\sout{2}}\\ \text{ =>AB= }\sqrt{18}\\ \text{ =>AB=3}\sqrt{2}\text{cm}\end{array} \end{gathered}$$

Question 17:

Fill in the blanks.

(i) The sum of the angles of a triangle is ...... .
(ii) The sum of any two sides of a triangle is always ...... than the third side.
(iii) In ∆ABC, if ∠A = 90°, then BC² = (......) + (......).
(iv) In ∆ABC, AB = AC and AD ⊥ BC, then BD = ...... .
(v) In the given figure, side BC of ∆ABC is produced to D and CE || BA. If ∠BAC = 50°
then ∠ACE = ...... .

Answer 17:

(i) The sum of the angles of a triangle is 180°.
(ii) The sum of any two sides of a triangle is always greater than the third side.
(iii) In ∆ABC, if ∠A = 90°, then:
BC² = (AB²) + (BC²)

$$\begin{gathered} \begin{array}{l}\text{ In }△\text{ABC, if }\angle \text{A }=90°, \mathrm{then} \mathrm{by} \mathrm{phythagoras} \mathrm{theorem}: \\ \end{array} \\ {\mathrm{BC}}^2={\text{AB}}^2+{\text{AC}}^2 \end{gathered}$$

(iv) In ∆ABC:
AB = AC
AD ⊥ BC
Then, BD = DC

$$\begin{gathered} \begin{array}{l}\text{BD}=\text{DC}\\ \end{array} \\ \text{This is because }\mathrm{in an isosceles triangle}, \mathrm{the} \mathrm{perpendicular dropped from} \mathrm{the vertex joining the equal sides}, bisects the base. \end{gathered}$$

(v) In the given figure, side BC of ∆ABC is produced to D and CE || BA.
If ∠BAC = 50°, then ∠ACE = 50°.

$\begin{array}{l}\text{AB II CE}\\ \text{ ∴ }\angle \text{BAC}=\angle \text{ACE}=50{°}^{} (\text{alternate angles)}\end{array}$

Question 18:

Write 'T' for true and 'F' for false

(i) If two parallel lines are cut by a transversal, then the alternate interior angles are equal.
(ii) If two lines intersect each other, then the vertically opposite angles are equal.
(iii) Each acute angle of an isosceles right triangle measures 60°.
(iv) A right triangle cannot have an obtuse angle.

Answer 18:

$$\begin{gathered} \begin{array}{l}\\ \left(\text{i}\right)\text{True}\\ \left(\text{ii}\right)\text{True}\\ \left(\text{iii}\right){\text{False. Each acute angle of an isoceles right triangle measures 45°.}}^{}\\ \left(\text{iv}\right)\text{True. This is because the sum of three angles of a triangle must be 180°. }\end{array} \\ \mathrm{So, if one of} {\text{the angles is 90°, the sum of the other two angles must also be 90°}}^{}\text{. } \\ \mathrm{So}, \text{none of the angles can be greater than 90°.} \end{gathered}$$