These AP 9th Class Maths Important Questions 7th Lesson Triangles will help students prepare well for the exams.

## AP State Syllabus 9th Class Maths 7th Lesson Important Questions and Answers Triangles

Question 1.

The interior angles of a triangle are (3x – 10)°, (3x + 10)°, (3x)°. Find the angles.

Solution:

Interior angles of a triangle are (3x- 10)°, (3x + 10)° and (3x)°

Sum of angles = (3x – 10)° + (3x + 10)° + (3x)° = 180°

9x = 180°

x = 20°

∴ Angles = 3 × 20 – 10 = 60 – 10 = 50°

= 3 × 20 + 10 = 60 + 10 = 70°

= 3 × 20° ± 60°

∴ Angles are 50°, 70° and 60°.

Question 2.

If ΔABC ≅ ΔPQR, express equal sides of triangles.

Solution:

If Δ ABC ≅ ΔPQR

⇒ AB = PQ, BC = QR, AC = PR

Question 3.

In ΔABC, the lines are drawn parallel to BC, CA and AB respectively through A, B, C intersecting at P, Q and R. Find the ratio of perimeter of ΔPQR and ΔABC

Solution:

From figure, AB // QP and BC // RQ.

So, in ABCQ parallelogram AB // CQ

Similarly, in BCAR, ABPC parallelograms BC = AQ and BC = RA.

Mid point of QR is ‘A’.

Similarly mid point of PR and PQ is B and C.

∴ AB = \(\frac{1}{2}\) PQ,

BC = \(\frac{1}{2}\) QR and CA = \(\frac{1}{2}\) PR.

∴ Perimeter of ΔPQR = PQ + QR + PR

= 2AB + 2BC + 2CA

= 2 (AB + BC + CA)

= 2(Perimeter of ΔABC)

∴ Ratio of perimeters of ΔPQR and ΔABC is 2 : 1.

Question 4.

In the adjacent figure, prove that ΔPQX ≅ ΔPRY.

Solution:

In ΔPQX, ΔPRY .

Sides QX = RY {Given}

Sides PX = PY {Given}

∠PXY = ∠PYX

180 – ∠PXY = 180° – ∠PYX

⇒ ∠PXQ = ∠PYR {Angle}

By S – A – S congruence criterion.

ΔPQX ≅ ΔPRY

Question 5.

In the given figure, AL is parallel to DC and E Is mid point of BC. Show that triangle EBL Is congruent to triangle ECD.

Solution:

From the adjacent figure AL // CD and E is the midpoint of BC.

i.e., BE = EC

from ΔBEL & ΔCED.

∠B = ∠C [ ∵ Alternate interior angles]

∠DEC = ∠BEL [ ∵ vertically opposite, angles]

∴ By A.S.A rule

Δ EBL ≅ Δ ECD.

Question 6.

ABC is a right angled triangle and M is the midpoint of the hypotenuse AB. A line segment MD, parallel to BC, passes through M and meets AC at D. Prove that (i) D is midpoint of AC, (ii) MD is perpendicular to AC.

Solution:

Given that ΔABC is a right angled triangle in which, M is the midpoint of AB. Also \(\overline{\mathrm{MD}}\) / / \(\overline{\mathrm{BC}}\).

∠C = 90° ⇒ BC⊥AC then MD JL AC [ ∵ BC // MD]

If a line is drawn parallel to one side and (divides the 2^{nd} side into two equal parts) passes through the midpoint of second side, it will also passes through the midpoint of 3^{rd} side, i.e., MD bisects AC.

i.e., D is the midpoint of AC.

Question 7.

In ΔPQR, PS⊥QR and ΔPQS ≅ ΔPRS.

PQ = 2x + 3, PR = 3y + 1, QS = x,

SR = y + 1. Find the area of ΔPQR.

Solution:

In ΔPQR, PS⊥QR, so ΔPRS ≅ ΔPQS From CPCT,

PQ = PR and QS = SR and 2x + 3 = 3y + 1 and x = y + 1

2x – 3y = -2 ………………… (1) and

x – y = 1 ………….. (2)

Substitute x = y + 1 in equation (1)

2(y + 1) – 3y = -2

2y + 2 – 3y = -2

-y = -4

⇒ y = 4

by y = 4 in x = y+ 1 ⇒ x = 5

∴ QR = x + y+ l= 5 + 4+ l = 10units

PQ = PR = 2(5) + 3 = 13 units

QS = x = 5 units.

∴ ΔPQS is a right triangle.

From pythagoras theorem,

PQ^{2} = QS^{2} + PS^{2}

PS^{2} = PQ^{2} – QS^{2}

PS^{2} = (13)^{3} – (5)^{2} = 144

∴ PS = 12 units

∴ Area of ΔPQR = \(\frac{1}{2}\) × QR × PS

= \(\frac{1}{2}\) × 10 × 12 = 60,sq.units.

Question 8.

In the given figure \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{EF}}\) at equal distances and AF is a transversal. \(\overline{\mathrm{GH}}\) is perpendicular to \(\overline{\mathrm{AB}}\). If AB = 4.5 cm, GH = 4 cm and FB = 8 cm, find the area of ΔGDF.

Solution:

In the given figure \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{EF}}\) at equal distances.

\(\overline{\mathrm{AF}}\) is a transversal and \(\overline{\mathrm{GH}}\) is perpendicular to \(\overline{\mathrm{AB}}\).

So, GH = 4 cm, AB = 4.5 cm, FB = 8 cm To find area of ΔGDF,

From ΔABF, D is mid point of \(\overline{\mathrm{BF}}\)

Similarly ‘G’ is mid point of \(\overline{\mathrm{AF}}\).

So BD ⊥ DF, AG = GF

Median divides a triangle into two equal triangles.

∴ Area of ΔABG = Area of ΔBGF

Similarly \(\overline{\mathrm{GD}}\) is median of ΔBGF.

Area of ΔABG = 2 × area of ΔDGF

Area of ΔDFG = \(\frac{1}{2}\) area of ΔABG

Area of ΔABG = \(\frac{1}{2}\) × 4.5 × 4 = 9 sq.cm

Area of ΔDGE = \(\frac{1}{2}\) × 9 = 4.5 sq.cm

Question 9.

In ABC, E and F are mid points of sides AB and AC respectively then prove that i) EF // BC and ii) EF = \(\frac{1}{2}\)BC.

Solution:

Given : B and F are mid points of AB and AC.

R.T.P. : i) EF // BC, ii) EF = \(\frac{1}{2}\) BC

Construction: ..

GC // AB, extend EF upto G.

Proof:

ΔAEF ACGF

∠AFE = ∠CFG (Vertically opposite angles)

AF = FC

∠EAF = ∠GCF (Alternate angles)

∴ ΔAEF = ΔCGF

∴ CG = BE and CG // BF (Construction)

∴ EBCG is a parallelogram.

Question 10.

In triangle ABC, D is a point on BC such that ΔABD and ΔACD are congruent. Find the values of x and y, if AB = 2x + 3, AC = 3y + 1, BD = x and

DC = y + 1.

Solution:

From the adjacent figure ΔABC, D is a point on BC.

Such that ΔABD ≅ ΔACD

AB = 2x + 3

AC = 3y + 1

BD = x

DC = Y + 1

Δ ABD = Δ ACD

⇒ AB = AC

⇒ 2x + 3 = 3y + 1

⇒ 2x – 3y + 2 = 0 …………….(1)

BD = CD ⇒ x = y + 1

⇒ x – y – 1 = 0 ………………….(2)

⇒ from (1) & (2)

y = 4

from (2) ⇒ x – 4 = 1

⇒ x = 1 + 4 = 5

∴ (x, y) = (5, 4)

Question 11.

In triangle ΔABC, D, E and F are mid points of sides AB’, BC and AC respectively. If we join points D, E and F, then show that triangle ABC is divided into four congruent triangles.

Solution:

Given : In ABC, D,E,F are the midpoints of sides AB, BC and AC respectively. .

R.T.P : ΔADF ≅ ΔDBE ≅ ΔDEF ≅ ΔCEF

Proof : D, E are the midpoints of \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BC}}\) of ΔABC respectively.

So by mid-point theorem, we know that DE // AC

Similarly DF//BC and EF // AB.

∴ ADEF, BEFD, CFDE are all parallelograms.

In a parallelogram ADEF, DF is the diagonal.

So, ΔADF ≅ ΔDEF …………….. (1)

[∵ Diagonal divides the parallelogram into two congruent triangles]

Similarly ΔBDE ≅ ΔDEF ………………… (2)

and ΔCEF ≅ ΔDEF …………….. (3)

From (1), (2) & (3) ⇒ ΔADF ≅ ΔDEF ≅ ΔBDE ≅ ΔCEF all the triangles are congruent to each other,

i.e., ΔABC divides into four congruent triangles by joining the mid points of the sides.