Table of Contents
NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 7 Maths are solved by experts in order to help students to obtain excellent marks in their annual examination. All the questions and answers that are present in the CBSE NCERT Books has been included in this page. We have provided all the Class 7 Maths NCERT Solutions with a detailed explanation i.e., we have solved all the question with step by step solutions in understandable language. So students having great knowledge over NCERT Solutions Class 7 Maths can easily make a grade in their board exams.
Chapter -6 The Triangle and its Properties
Ex 6.1 Class 7 Maths Question 1.
Solution:
Ex 6.1 Class 7 Maths Question 2.
Draw rough sketches for the following :
(a) In ∆ABC, BE is a median.
(b) in ∆PQR, PQ and PR are altitudes of the triangle.
(c) In ∆XYZ, YL is an altitude In the exterior of the triangle.
Solution:
(a) Rough sketch of median BE of ∆ABC is as shown.
(b) Rough sketch of altitudes PQ and PR of ∆PQR is as shown.
(c) Rough sketch of an exterior altitude YL of ∆XYZ is as shown.
Ex 6.1 Class 7 Maths Question 3.
Verify by drawing a diagram if ‘the median and altitude of an isosceles triangle can be same.
Solution:
Draw a line segment BC. By paper folding, locate the perpendicular bisector of BC. The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and AC. The triangle thus obtained is an isosceles ∆ABC in which AB = AC.
Since D is the mid-point of BC, so AD is its median. Also, AD is the perpendicular bisector of BC. So, AD is the altitude of ∆ABC.
Thus, it is verified that the median and altitude of an isosceles triangle are the same.
Ex 6.2 Class 7 Maths Question 1.
Find the value of the unknown exterior angle x in the following diagrams :
Solution:
Since, in a triangle an exterior angle is equal to the sum of the two interior opposite angles, therefore,
- x = 50°+ 70° = 120°
- x = 65°+ 45° = 110°
- x = 30°+ 40°= 70°
- x = 60° + 60° = 120c
- x = 50° + 50° = 100c
- x = 30°+ 60° = 90°
Ex 6.2 Class 7 Maths Question 2.
Find the value of the unknown interior angle x in the following figures :
Solution:
We know that in a triangle, an exterior angle is equal to the sum of the two interior opposite angles. Therefore,
Ex 6.3 Class 7 Maths Question 1.
Find the value of the unknown x in the following diagrams :
Solution:
Ex 6.3 Class 7 Maths Question 2.
Find the values of the unknowns x and y in the following diagrams :
Solution:
Ex 6.4 Class 7 Maths Question 1.
Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
Solution:
(i) Since, 2 + 3 > 5
So the given side lengths cannot form a triangle.
(ii) We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3
i. e., the sum of any two sides is greater than the third side.
So, these side lengths form a triangle.
(iii) We have, 6 + 3 > 2, 3 + 2 Undefined control sequence \ngtr 6
So, the given side lengths cannot form a triangle.
Ex 6.4 Class 7 Maths Question 2.
Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR ?
(iii) OR + OP > RP ?
Solution:
(i) Yes, OP + OQ > PQ because on joining OP and OQ, we get a ∆OPQ and in a triangle, sum of the lengths of any two sides is always greater than the third side.
(ii) Yes, OQ + OR > QR, because on joining OQ and
OR, we get a ∆OQR and in a triangle, sum of the length of any two sides is always greater than the third side.
(iii) Yes, OR + OP > RP, because on joining OR and OP, we get a ∆OPR and in a triangle, sum of the lengths of any two sides is always greater than the third side.
Ex 6.4 Class 7 Maths Question 3.
AM is median of a triangle ABC. Is AB + BC + CA > 2AM?
(Consider the sides of triangles ∆ABM and ∆AMC.)
Solution:
Using triangle inequality property in triangles ABM and AMC, we have
AB + BM > AM …(1) and, AC + MC > AM …(2)
Adding (1) and (2) on both sides, we get
AB + (BM + MC) + AC > AM + AM
⇒ AB + BC + AC > 2AM
Ex 6.4 Class 7 Maths Question 4.
ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?
Solution:
Ex 6.4 Class 7 Maths Question 5.
ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?
Solution:
Let ABCD be a quadrilateral and its diagonals AC and BD intersect at O. Using triangle inequality property, we have
In ∆OAB
OA + OB > AB ……(1)
Ex 6.4 Class 7 Maths Question 6.
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Solution:
Let x cm be the length of the third side.
Thus, 12 + 15 > x, x + 12 > 15 and x + 15 > 12
⇒ 27 > x, x > 3 and x > -3
The numbers between 3 and 27 satisfy these.
∴ The length of the third side could be any length between 3 cm and 27 cm.
Ex 6.5 Class 7 Maths Question 1.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR? = 24 cm, find QR.
Solution:
Ex 6.5 Class 7 Maths Question 2.
ABC is a triangle right-angled atC. If AB = 25 cm and AC = 7cm, find BC.
Solution:
Ex 6.5 Class 7 Maths Question 3.
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:
Hence, the distance of the foot of the ladder from the wall is 9 m.
Ex 6.5 Class 7 Maths Question 4.
Which of the following can be the sides of a right triangle?
(i) 2.5 em, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.
Solution:
Thus, the given sides form a right-triangle and the right-angle is opposite to the side of length 2.5 cm.
Ex 6.5 Class 7 Maths Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Solution:
Let ACB be the tree before it broke at the point C. and let its top A touch the ground at A‘ after it broke. Then, ∆A‘BC is a right triangle, right-angled at B such that A’ B = 12 m, BC = 5 m. By Pythagoras theorem, we have
Ex 6.5 Class 7 Maths Question 6.
Angles Q and ii of a APQR are 25° and 65°. Write which of the following is true :
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
Solution:
Ex 6.5 Class 7 Maths Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Solution:
Let ABCD be a rectangle such that AB = 40 m and AC = 41 m.
In right-angled ∆ABC, right-angled at B, by Pythagoras theorem, we have BC2 = AC2 – AB2
Ex 6.5 Class 7 Maths Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:
Let ABCD be the rhombus such that AC = 30 cm and BD = 16 cm.
We know that the diagonals of a rhombus bisect each other at right angles.
Related
Discover more from EduGrown School
Subscribe to get the latest posts sent to your email.