NCERT Solution class 7 Mathematics EXERCISE- 6.5
CHAPTER-6
EXERCISE – 6.5
Q1. PQR is a triangle , right angled at P. If PQ = 10 cm and PR = 24 cm , find QR.
SOLUTION:-
Let us draw a rough sketch of right – angled triangle
By the rule of Pythagoras theorem,
(QR)2= (RP)2 + (PQ)2
(QR)2= (24)2 + (10)2
(QR)2= 576 + 100
(QR)2= 676
QR = (676)2
QR = 26 cm.
Q2. ABC is a triangle , right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
SOLUTION:-
By the rule of Pythagoras theorem,
AB2 = AC2 + BC2
252 = 72 + BC2
625 = 49 + BC2
625 – 49 = BC2
576 = BC2
(576)2 = BC
24 = BC
BC = 24 cm.
Q3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
SOLUTION:-
By the rule of Pythagoras theorem,
(15)2 = (12)2 + a2
225= 144 + a2
225 -144 = a2
81 = a2
(81)2 = a
9 = a
a = 9 cm.
Q4. Which of the following can be the sides of a right triangle?
(i) 2.5 cm , 6.5 cm , 6 cm.
(ii) 2 cm , 2 cm , 5 cm.
(iii) 1.5 cm , 2 cm , 2.5 cm.
In case of right – angled triangles, identify the right angles.
SOLUTION:-
(i) This triangle is right – angled triangle.
(ii) This triangle is not right – angled triangle.
(iii) This triangle is right – angled triangle.
Q5. A tree is broken at a height of 5m from the ground and its top touches the ground at distance of 12m from the base of the tree. Find the original height of the tree.
SOLUTION:-
Let ABC is the triangle and B is the mid – point where tree is broken at the height 5 m from the ground.
Tree top touches the ground at a distance of AC = 12 m from the base of the tree,
By observing the figure we came to conclude that right angle triangle is formed at A.
From the rule of Pythagoras theorem,
BC2 = AB2 + AC2
BC2 = 52 + 122
BC2 = 25 + 144
BC2 = 169
BC = (169)2
BC = 13
Then, the original height of the tree = AB + BC
= 5 + 13
= 18 m
Q6. Angles Q and R of a ΔPQR are 250and 650 Write which of the following is true:
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
SOLUTION:-
Given that Angle B = 350, Angle C = 550
Then, Angle A = ?
We know that sum of the three interior angles of triangle is equal to 1800.
= Angle PQR + Angle QRP + Angle RPQ = 1800
= 250 + 650 + Angle RPQ = 1800
= 900 + Angle RPQ = 1800
= Angle RPQ = 1800 – 900
= Angle RPQ = 900
Also, we know that side opposite to the right angle is the hypotenuse.
QR2 = PQ2 + PR2
Hence, (ii) is true
Q7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
SOLUTION:-
Let ABCD be the rectangular plot.
Then, AB = 40 cm and AC = 41 cm.
BC= ?
According to Pythagoras theorem,
From right angle triangle ABC, we have:
= AC2 = AB2 + BC2
= 412 = 402 + BC2
= BC2 = 412 – 402
= BC2 = 1681 – 1600
= BC2 = 81
= BC = 812
= BC = 9 cm
Hence, the perimeter of the rectangle plot = 2(length + breadth)
Where , length = 40 cm,
Breadth = 9 cm
Then,
= 2(40+9)
= 2(49)
= 98 cm.
Q8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
SOLUTION:-
Let PQRS be a rhombus , all sides of rhombus has equal length and its diagonal PR and SQ are intersecting each other at point O. Diagonals in rhombus bisect each other at 900
So, PO = (PR/2)
= 16/2
= 8 cm
And, SO = (SQ/2)
= 30/2
= 15 cm
Then, consider the triangle POS and apply the Pythagoras theorem,
PS2 = PO2 + SO2
PS2 = 82 + 152
PS2 = 64 +225
PS2 = 289
PS = 2892
PS = 17 cm
Hence , the length of side of rhombus is 17 cm.
Now,
Perimeter of rhombus = 4 x side of the rhombus
= 4 x 17
= 68 cm
Perimeter of rhombus is 68 cm.