Exercise 11D
Question 1
(a) What will be the other angles of a right angled isosceles triangle ?
Sol :
Figure to be added
In right angled isosceles triangle ABC
∠B=90°
AB=BC
∠A=∠C
∴90°+∠A+∠C=180°
90°+2∠A=180°
∠A$=\frac{90}{2}$
∴∠A=45°
Ans : The other triangle of a right angled isosceles triangle is 45°
(b) Can you draw an obtuse angled isosceles triangle?
Figure to be added
Ans : Yes, I can draw an obtuse angled isosceles triangle
Question 2
(a) The vertical angle of an isosceles triangle is 110° .What must be the size of each of its base angles ?
Sol :
Let, the size of each of base angles=x
∴2x+100°=180°
2x=180°-110°
2x=70°
$x=\frac{70}{2}$
∴x=35°
Ans : The size of each base angle is 35°
(b) What is the size of each exterior angle of an equilateral triangle ?
Sol :
In a equilateral triangle each angle is 60°
∴Each exterior angle of an equilateral triangle is 180°-60°=120°
Question 3
ΔABC is isosceles with AB=AC , if ∠A=80°. What is the measure of ∠B?
Figure to be added
∠B=∠C
∴80°+∠B+∠C=180°
80°+2∠B=180°
2∠B=180°-80°
$=\frac{100}{2}$
∠B=50°
Question 4
In ΔABC, ∠A=∠B=50°. Name the pair of sides which are equal
Figure to be added
Ans : AC and BC are equal
Question 5
In the figure given alongside , AN=AC, ∠BAC=52°, ∠ACK=84° and BCK is a straight line. Prove that NB=NC
Figure to be added
In ΔANC, AN=AC
∴∠ANC=∠ACN
∴52°+∠ANC+∠ACN=180°
2∠ANC=180°-52°
∠ANC$=\frac{128}{2}$
∴∠ANC=64°
In ΔNBC,
∴∠BNC=180°-64°=116°
∴∠NCB=180°-(64°+84°)
=180°-148°=32°
∴∠NCB=180°-(116°+32°)
=180°-32°
∠NCB=∠NBC
∴ΔNBC is an isosceles triangle
∴NB=NC
Question 6
In the figure , AB=AC. Prove that BD=BC
Figure to be added
In ΔABC, ∠ADB=180°-(40°+30°)
=180°-70°=110°
∴∠BDC=180°-110°=70°
In ΔABC, AB=AC
∴∠ABC=∠ACB
40°+∠ABC+∠ACB=180°
40°+2∠ABC=180°
2∠ABC=180°-40°
∠ABC$=\frac{140}{2}$
∴∠ABC=∠ACB=70°
∴∠DBC=70°-30°=40°
In ΔBDC,
∴∠BDC=∠DCB
∴ΔBDC is an isosceles triangle
∴BD=BC
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