RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

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Class 7 : RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

1. ∠CBX is an exterior angle of ΔABC at B. Name

(i) the interior adjacent angle

(ii) the interior opposite angles to exterior ∠CBX

Also, name the interior opposite angles to an exterior angle at A.

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

(i) ∠ABC

(ii) ∠BAC and ∠ACB

Also the interior angles opposite to exterior are ∠ABC and ∠ACB

2. In the fig, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

In ΔABC, ∠A = 50° and ∠B = 55°

Because of the angle sum property of the triangle, we can say that

∠A+∠B+∠C = 180°

50° +55° +∠C = 180°

Or

∠C=75°

∠ACB = 75°

∠ACX = 180°−∠ACB = 180°−75° = 105°

3. In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

We know that the sum of interior opposite angles is equal to the exterior angle.

Hence, for the given triangle, we can say that :

∠ABC+ ∠BAC = ∠BCO

55° + ∠BAC = 95°

Or,

∠BAC = 95°– 95°

= ∠BAC = 40°

We also know that the sum of all angles of a triangle is 180°.

Hence, for the given ΔABC, we can say that :

∠ABC + ∠BAC + ∠BCA = 180°

55° + 40° + ∠BCA = 180°

Or,

∠BCA =180° –95°

= ∠BCA = 85°

4. One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?

Solution.

Let us assume that A and B are the two interior opposite angles.

We know that ∠A is equal to ∠B.

We also know that the sum of interior opposite angles is equal to the exterior angle.

Hence, we can say that :

∠A + ∠B = 80°

Or,

∠A +∠A = 80° (∠A= ∠B)

2∠A = 80°

∠A = 40°2 =40°

∠A= ∠B = 40°

Thus, each of the required angles is of 40°.

5. The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

In the given figure, ∠ABE and ∠ABC form a linear pair.

∠ABE + ∠ABC =180°

∠ABC = 180°– 136°

∠ABC = 44°

We can also see that ∠ACD and ∠ACB form a linear pair.

∠ACD + ∠ACB = 180°

∠AUB= 180°– 104°

∠ACB = 76°

We know that the sum of interior opposite angles is equal to the exterior angle.

Therefore, we can say that :

∠BAC + ∠ABC =104°

∠BAC = 104°– 44° = 60°

Thus,

∠ACE = 76°

and

∠BAC = 60°

6. In Fig, the sides BC, CA and BA of a ΔABC have been produced to D, E and F respectively. If ∠ACD=105° and ∠EAF=45° ; find all the angles of the ΔABC

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

In a ΔABC, ∠BAC and ∠EAF are vertically opposite angles.

Hence, we can say that :

∠BAC= ∠EAF = 45°

Considering the exterior angle property, we can say that :

∠BAC + ∠ABC = ∠ACD = 105°

∠ABC = 105°– 45° = 60°

Because of the angle sum property of the triangle, we can say that :

∠ABC+ ∠ACS +∠BAC = 180°

∠ACB = 75°

Therefore, the angles are 45°, 65° and 75°.

7. In Fig, AC perpendicular to CE and C ∠A:∠B:∠C=3:2:1. Find the value of ∠ECD.

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

In the given triangle, the angles are in the ratio 3 : 2 : 1.

Let the angles of the triangle be 3x, 2x and x.

Because of the angle sum property of the triangle, we can say that :

3x+2x+x = 180°

6x = 180°

Or,

x = 30° … (i)

Also, ∠ACB + ∠ACE + ∠ECD = 180°

x+ 90° + ∠ECD = 180° (∠ACE = 90° )

∠ECD= 60° [From (i)]

8 A student when asked to measure two exterior angles of ΔABC observed that the exterior angles at A and B are of 103° and 74° respectively. Is this possible? Why or why not?

Here,

Solution.

Internal angle at A+ External angle at A = 180°

Internal angle at A + 103° =180°

Internal angle at A = 77°

Internal angle at B + External angle at B = 180°

Internal angle at B + 74° = 180°

Internal angle at B = 106°

Sum of internal angles at A and B = 77° + 106° =183°

It means that the sum of internal angles at A and B is greater than 180°, which cannot be possible.

9. In Fig, AD and CF are respectively perpendiculars to sides BC and AB of ΔABC. If ∠FCD=50°, find ∠BAD

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

We know that the sum of all angles of a triangle is 180°

Therefore, for the given ΔFCB, we can say that :

∠FCB+ ∠CBF+ ∠BFC= 180°

50° + ∠CBF + 90°= 180°

Or,

∠CBF = 180° –50°– 90° = 40° … (i)

Using the above rule for ΔABD, we can say that :

∠ABD + ∠BDA + ∠BAD = 180°

∠BAD = 180° –90°– 40° = 50° [from (i)]

10. In Fig, measures of some angles are indicated. Find the value of x.

RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev

Solution.

Here,

∠AED + 120° = 180° (Linear pair)

∠AED = 180°– 120° = 60°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ΔADE, we can say that :

∠ADE + ∠AED + ∠DAE = 180°

60°+ ∠ADE + 30° =180°

Or,

∠ADE =180°– 60°– 30° = 90°

From the given figure, we can also say that :

∠FDC + 90° = 180° (Linear pair)

∠FDC = 180°– 90° = 90°

Using the above rule for ΔCDF, we can say that :

∠CDF + ∠DCF + ∠DFC = 180°

90° + ∠DCF + 60° =180°

∠DCF = 180°−60°−90°= 30°

Also,

∠DCF + x = 180° (Linear pair)

30° + x = 180°

Or,

x = 180°– 30° = 150°

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