4 Mark Question

Question 14 Marks

In the adjoining figure, point S is any point on side QR of ∆PQR. Prove that: PQ + QR + RP > 2PS

Answer

Proof:
In ∆PQS,
PQ + QS > PS …..(i) [Sum of any two sides of a triangle is greater than the third side]
Similarly, in ∆PSR,
PR + SR > PS …(ii) [Sum of any two sides of a triangle is greater than the third side]
∴ PQ + QS + PR + SR > PS + PS
∴ PQ + QS + SR + PR > 2PS
∴ PQ + QR + PR > 2PS [Q-S-R]

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Question 24 Marks

We have learnt that if two triangles are equiangular then their sides are in proportion. What do you think if two quadrilaterals are equiangular? Are their sides in proportion? Draw different figures and verify. Verify the same for other polygons.

Answer

If two quadrilaterals are equiangular then their sides will not necessarily be in proportion.
Case 1: The two quadrilaterals are of the same type.

Consider squares ABCD and PQRS.
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S
$\frac{ AB }{ PQ }=\frac{ BC }{ QR }=\frac{ CD }{ RS }=\frac{ AD }{ PS }$
Case 2: The two quadrilaterals are of different types.

Consider square ABCD and rectangle STUV.
∠A = ∠S, ∠B = ∠T, ∠C = ∠U, ∠D = ∠V
Now, $\frac{ AB }{ ST }=\frac{ CD }{ UV }$ and $\frac{ BC }{ TU }=\frac{ AD }{ SV }$ But $\frac{ AB }{ ST } \neq \frac{ BC }{ TU }$

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Question 34 Marks

Find the values of x and y using the information shown in the given figure. Find the measures of ∠ABD and ∠ACD.

Answer

i. ∠ACB = 50° [Given]
In ∆ABC, seg AC ≅ seg AB [Given]
∴ ∠ABC ≅ ∠ACB [Isosceles triangle theorem]
∴ x = 50°

ii. ∠DBC = 60° [Given]
In ABDC, seg BD ≅ seg DC [Given]
∴ ∠DCB ≅ ∠DBC [Isosceles triangle theorem]
∴ y = 60°

iii. ∠ABD = ∠ABC + ∠DBC [Angle addition property]
= 50° + 60°
∴ ∠ABD = 110°

iv. ∠ACD = ∠ACB + ∠DCB [Angle addition property]
= 50° + 60°
∴ ∠ACD = 110°
∴ x = 50°, y = 60°,
∠ABD = 110°, ∠ACD = 110°

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Question 44 Marks

In the given figure, bisectors of $\angle B$ and $\angle C$ of $\triangle A B C$ intersect at point P. Prove that $\angle BPC =90^{\circ}+\frac{1}{2} \angle B A C$.
Complete the proof by filling in the blanks.

Answer

Proof:
In ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180° …[Sum of the measures of the angles of a triangle is 180°]
$\therefore \angle BAC +-\angle ABC +\angle ACB =180 \ldots$ [Multiplying each term by $\frac{1}{2}$ ]
∴ ∠BAC + ∠PBC + ∠PCB = 90°
∴ ∠PBC + ∠PCB = 90° – 1 ∠BAC ………(i)
In∆BPC,
∠BPC + ∠PBC + ∠PCB = 180° …….[Sum of measures of angles of a triangle]
$\therefore \angle B P C+90^{\circ}-\frac{1}{2} \angle B A C=180^{\circ} \ldots . .[\text { From (i)] }$
$\therefore \angle B P C=180^{\circ}-90^{\circ} \frac{1}{2} \angle B A C$
$=180^{\circ}-90^{\circ}+\frac{1}{2} \angle B A C$
$=90^{\circ}+\frac{1}{2} \angle B A C$

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Question 54 Marks

In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.

Answer

i. ∠NET = 100° and ∠EMR = 140°
∠EMN + ∠EMR = 180°
∴ z +140° =180°
∴ z = 180° -140°
∴ z = 40°

ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair]
∴ 100° + y = 180°
∴ y = 180° – 100°
∴ y = 80°

iii. In ∆ENM,
∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ x +80°+ 40°= 180°
∴ x = 180° – 80° – 40°
∴ x = 60°
∴ x = 60°, = 80°, z = 40°

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Question 64 Marks

The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.

Answer

Let the measure of the smallest angle be x°.
One of the angles is twice the measure of the smallest angle.
∴ Measure of that angle = 2x°
Another angle is thrice the measure of the smallest angle.
∴ Measure of that angle = 3x°
∴ The measures of the remaining two angles are 2x° and 3x°.
Now, x° + 2x° + 3x° = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 6x = 180
∴ x = 180
$\therefore x=\frac{180}{6}$
∴ x° = 30°
The measures of the remaining angles are 2x° = 2 x 30° = 60°
3x° = 3 x 30° = 90°
The measures of the three angles of the triangle are 30°, 60° and 90°.

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