Ncert – Class 10 –Triangles-Hots Question

Question 1

A street light bulb is fixed on a pole 6m above the level of the street. If a woman of height 1.5m
casts a shadow of 3m. then find how far she is away from the base of the pole ?

Solution

X = 8m

Question 2

An aeroplane leaves an Airpot and flies due North at speed 1000 km per hour. At the same time
another aeroplane leaves the same airpot and flies due west at a speed of 1200km per hour. How
far apart will be the two planes after one and half hours ?

Solution

AB = 30061 KM

Question 3

A vertical pole of 6m casts a shadow 4m long on the ground and at the same time a tower
casts a shadow 28m long. Find the height of the tower

Solution

Length of the vertical pole = 6 m
Shadow of the pole = 4 m
Let the height of the tower be h m.
5
Length of the shadow of the tower = 28 m
In ΔABC and ΔDFE,
∠C = ∠E (angle of elevation)
∠B = ∠F = 90°
By AA similarity criterion,
ΔABC ~ ΔDFE
We know that the corresponding sides of two similar triangles are proportional.
AB/DF = BC/EF
6/h = 4/28
h = (6 ×28)/4
h = 6 × 7
h = 42
Hence, the height of the tower = 42 m

Question 4

In a ∆ PQR, PR^2 –PQ^2 = QR^2 and M is a point on side PR such that QM ⊥ PR. Prove that
QM² = PM × MR

Solution

Given:
In ∆ PQR, PR²-PQ²= QR² & QM ⊥ PR
To Prove: QM² = PM × MR
Proof:
Since, PR² – PQ²= QR²
PR² = PQ² + QR²
So, ∆ PQR is a right-angled triangle at Q.
In ∆ QMR & ∆PMQ
∠QMR = ∠PMQ [ Each 90°]
∠MQR = ∠QPM [each equal to (90°- ∠R)]
∆ QMR ~ ∆PMQ [ by AA similarity criterion]
By the property of an area of similar triangles,
ar(∆ QMR ) / ar(∆PMQ)= QM²/PM²
1/2× MR × QM / ½ × PM ×QM = QM²/PM²
[ Area of triangle= ½ base × height]
MR / PM = QM²/PM²
QM² × PM = PM² × MR
QM² =( PM² × MR)/ PM

QM² = PM × MR.

Question 5

For going to the city B from the city A, there is a route via city C such that perpendicular,
AC⊥CB, AC = 2 x km as well as CB = 2 (x + 7) km. It is proposed to build a 26 km highway that directly connects the two cities A and B. Find out how much distance will be saved in reaching city B from city A after the construction of the highway

Solution

distance saved = 34 – 26 = 8 km.

Question 6

If  ABC   DEF, AB = 4cm , DE = 6cm, EF = 9cm and FD = 12cm. Find the perimeter of the  ABC.

Solution

𝑥^2 = 144
X =12

Question 7

Are of two similar triangles are 36cm2 and 100cm2. If the length of a side of the larger triangle is
20cm. Find the length of the corresponding side of the smaller triangle

Solution

BC =6cm
Ac = 8cm
Perimeter of  ABC = AB + BC + CA
= 4 + 6 + 8 = 18cm

Question 8

ABCD is a trapezium in which AB||DC and P and Q are points on AD
and BC, respectively such that PQ||DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD

Solution

Given, PD = 18 cm
BQ = 35 cm
QC = 15cm
Also, AB||PQ
In △ABD,
Given, AB||PQ
So, PO||AB
By BPT, DP/AP = OD/OB ————– (1)
In △BDC,
Given, DC||PQ
So, OQ||DC
By BPT, OB/OD = BQ/QC
On rearranging,
OD/OB = QC/BQ ————————— (2)
Equating (1) and (2),
DP/AP = QC/BQ
So, 18/AP = 15/35
AP(15) = 35(18)
4
AP = 7(18)/3
AP = 42 cm
We know, AD = DP + AP
AD = 18 + 42
AD = 60 cm
Therefore, the length of AD is 60 cm

Question 9

A high tower casts a shadow 24m long at a certain time and at the same time a telephone p0le
casts a shadow of 16m long. Find the height of the telephone pole ?

Solution

H = 10m

Question 10

Two poles of height 6m and 11m stand on a plane ground. If the distance between the feet of the
poles is 12m. Find the distance between their tops ?

Solution

AC = 13cm