anyone explain properties of triangles (class 7)
Dont copy from google plz
Answers
Answer:
Solutions for Class 7 Math Chapter 15 Properties Of Triangles are provided here with simple step-by-step explanations. These solutions for Properties Of Triangles are extremely popular among Class 7 students for Math Properties Of Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs_aggarwal_(2018) Book of Class 7 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs_aggarwal_(2018) Solutions. All Rs_aggarwal_(2018) Solutions for class Class 7 Math are prepared by experts and are 100% accurate.
Page No 183:
Question 1:
In a ∆ABC, if ∠A = 72° and ∠B = 63°, find ∠C.
ANSWER:
Sum of the angles of a triangle is 180°.
∴∠A+∠B+∠C = 180°72° + 63° +∠C = 180°∠C = 45°
Hence, ∠C measures 45°.
Page No 183:
Question 2:
In a ∆DEF, if ∠E = 105° and ∠F = 40°, find ∠D.
ANSWER:
Sum of the angles of any triangle is 180°.
In ∆DEF:
∠D+∠E+∠F=180°∠D+105°+40°=180°or ∠D=180°−(105°+40°)or ∠D=35°
Page No 183:
Question 3:
In a ∆XYZ, if ∠X = 90° and ∠Z = 48°, find ∠Y.
ANSWER:
Sum of the angles of any triangle is 180°.
In ∆XYZ:
∠X+∠Y+∠Z=180°90°+∠Y+48°=180°=>∠Y=180°−138°=42°
Page No 183:
Question 4:
Find the angles of a triangle which are in the ratio 4 : 3 : 2.
ANSWER:
Suppose the angles of the triangle are (4x)o, (3x)o and (2x)o.
Sum of the angles of any triangle is 180o.
∴ 4x + 3x + 2x = 180
9x = 180
x = 20
Therefore, the angles of the triangle are (4×20)°, (3×20)° and ( 2×20)°, i.e . 80°, 60° and 40°.
Page No 183:
Question 5:
One of the acute angles of a right triangle is 36°. find the other.
ANSWER:
Sum of the angles of a triangle is 180°.
Suppose the other angle measures x.
It is a right angle triangle. Hence, one of the angle is 90°.
∴ 36° + 90° +x = 180°x= 54°
Hence, the other angle measures 54°.
Page No 183:
Question 6:
The acute angles of a right triangle are in the ratio 2 : 1. Find each of these angles.
ANSWER:
Suppose the acute angles are (2x)° and (x)°
Sum of the angles of any triangle is 180°
∴ 2x+x+ 90 = 180
⇒(3x) = 180-90
⇒(3x) = 90
⇒ x = 30
So, the angles measure (2×30)° and 30°i.e. 60° and 30°
Page No 183:
Question 7:
One of the angles of a triangle is 100° and the other two angles are equal. Find each of the equal angles.
ANSWER:
The other two angles are equal. Let one of these angles be x°.
Sum of angles of any triangle is 180°.
∴ x + x+ 100 = 180
2x = 80
x = 40
Hence, the equal angles of the triangle are 40° each.
Page No 184:
Question 8:
Each of the two equal angles of an isosceles triangle is twice the third angle. Find the angles of the triangle.
ANSWER:
Suppose the third angle of the isosceles triangle is xo.
Then, the two equal angles are (2x)o and (2x)o.
Sum of the angles of any triangle is 180o.
∴ 2x +2x+ x= 180
5x = 180
x = 36
Hence, the angles of the triangle are 36°, (2×36)° and (2×36)°, i.e. 36°, 72°and 72°.
Page No 184:
Question 9:
If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.
ANSWER:
Suppose the angles are ∠A, ∠B and∠ C.Given: ∠A = ∠B +∠CAlso, ∠A +∠B+∠C = 180°∴ ∠A+∠A= 180°⇒2∠A = 180°⇒ ∠A=90° (Sum of the angles of a triangle is 180°)
Hence, the triangle ABC is right angled at ∠A.
Page No 184:
Question 10:
In a ∆ABC, if 2∠A = 3∠B = 6∠C, calculate ∠A, ∠B and ∠C.
ANSWER:
Suppose: 2∠A = 3∠B = 6∠C = x°
Then, ∠A = (x2)∘
∠B =(x3)∘and ∠C =(x6)°
Sum of the angles of any triangle is 180°.
∠A +∠B +∠C = 180°
⇒x2+x3+x6= 180°⇒3x+2x+x6=180°⇒6x6=180°⇒x=180
∴ ∠A=(1802)∘=90°
∠B =(1803)∘=60∘∠C =(1806)∘=30°
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