Practice Test: Triangles - Grade 10 MCQ

Since O is the point of intersection of two equal chords AB and CD such that OB = OD,
As chords are equal and OB = OD, so AO will also be equal to OC
Also ∠AOC = ∠DOB = 45⁰
Now in triangles OAC and ODB
AO/OB = CO/OD
And ∠AOC = ∠DOB = 45⁰
So triangles are isosceles and similar.

According to given question

The far end of shadow is represented by point A,
Therefore we need to Find AC
By Pythagoras theorem,
(18)² + (9.6)² = (AC)²
⇒ AC² = 416.16
⇒ AC = 20.4 m (approx)

Since, PS is the angle bisector of angle QPR
So, by angle bisector theorem,
QS/SR = PQ/PR
⇒ 3/SR = 6/8
⇒ SR = (3 X 8)/6 cm = 4 cm

Let AC be the ladder of length 5m and BC = 4m be the height of the wall where ladder is placed. If the foot of the ladder is moved 1.6m towards the wall i.e. AD = 1.6 m, then the ladder is slided upward to position E i.e. CE = x m.

In right triangle ABC
AC² = AB² + BC²
⇒5² = AB² + 4²
⇒ AB = 3m
⇒ DB = AB – AD = 3 – 1.6 = 1.4m
In right angled ΔEBD
ED² = EB² + BD²
⇒ 5² = EB² + (1.4)²
⇒ EB = 4.8m
EC = EB – BC = 4.8 – 4 = 0.8m

According to given Question
ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of larger triangle)²
ar(Larger Triangle)/48 = (3/2)²
ar(Larger Triangle) = (9 x 48 )/4
ar(Larger Triangle) = 108 cm²

In triangle ABC, we have DE || BC
∴ AD/DB = AE/EC (By Thale’s Theorem)
⇒ x/x – 2 = (x + 2)/(x – 1)
⇒ x (x – 1) = (x – 2)(x + 2)
⇒ x² – x = x² – 4
⇒ x = 4

According to question,
ΔABC ~ ΔDEF,
AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm,
Therefore,
AB/DE = BC/EF = AC/DF
4/6 = BC/9 = AC/12
⇒ 4/6 = BC/9
⇒ BC = 6 cm
And
4/6 = AC/12
⇒ AC = 8 cm
Perimeter = AB + BC + CA
= 4 + 6 + 8
= 18 cm

The altitude divides the opposite side into two equal parts,
Therefore, BD = DC = 4 cm

In triangle ABD
AB² = AD² + BD²
8² = AD² + 4²
AD² = 64 – 16
AD² = 48
AD = 4√3 cm

In triangle ACB and ADC
∠A=∠A
∠ACB = ∠CDA
Therefore triangle ACB and ADC are similar,
Hence
AC/AD = AB/AC
AC² = AD X AB
8² = 3 x AB
⇒ AB = 64/3
This implies,
BD = 64/3 – AD
⇒ BD = 55/3

In ΔALD, we have
BP || AD
∴ LB/BA = LP/PD
⇒ BL/AB = PL/DP
⇒ BL/DC = PL/DP [∵ AB = DC
⇒ DP/PL = DC/BL