Class 9 Maths: Triangles – Extra Questions with Answers

Ans: A

Solution:
2x + 5 = 11
2x = 11 – 5
2x = 6
x = 6 / 2
x = 3
Answer: 3

Ans: D

Solution: The exterior angle is equal to the sum of the two interior opposite angles. Therefore, the other interior angle is $120^\circ – 50^\circ = 70^\circ$. $70^\circ = K \times 10^\circ$, so $K=7$.
Answer: 7

Ans: C

Solution: The altitude bisects the base, forming two right triangles. Using the Pythagorean theorem, half the base = √(17² – 8²) = √(289 – 64) = √225 = 15 cm. Therefore, the base = 2 * 15 = 30 cm.
Answer: 30 cm

Ans: B

Solution:
Let the two angles be $x$ and $y$. We have $x+y=122$ and $x-y=30$.
Adding the two equations, $2x=152$, so $x=76$.
Then $y=122-76=46$.
The third angle is $180-122=58$.
Answer: $76^\circ, 46^\circ, 58^\circ$

Ans: A

Solution:
$y^2 + 64 = 100$
$y^2 = 100 – 64$
$y^2 = 36$
$y = \sqrt{36}$
$y = 6$

Answer: 6

Ans: B

Solution:
Use the Pythagorean theorem to find the other side: a^2 + b^2 = c^2. 8^2 + b^2 = 17^2, 64 + b^2 = 289, b^2 = 225, b = 15. Perimeter = 2(length + width) = 2(8 + 15) = 2(23) = 46.
Answer: 46 meters

Ans: C

Solution:
Using the 3D Pythagorean theorem, the longest diagonal is sqrt(3^2 + 6^2 + 4^2) = sqrt(9 + 36 + 16) = sqrt(61).
Answer: sqrt(61) m

Ans: D

Solution:
We know that $\angle A + \angle B + \angle C = 180^\circ$.
From the given information, $\angle A + \angle B = 130^\circ$. Substituting this into the sum of angles of a triangle, we have $130^\circ + \angle C = 180^\circ$, which means $\angle C = 180^\circ – 130^\circ = 50^\circ$.
We are also given $\angle A + \angle C = 110^\circ$. Since we found that $\angle C = 50^\circ$, we can substitute this into the equation, so $\angle A + 50^\circ = 110^\circ$. This gives us $\angle A = 110^\circ – 50^\circ = 60^\circ$.
Now, we can find $\angle B$. We know that $\angle A + \angle B = 130^\circ$, so $60^\circ + \angle B = 130^\circ$. This implies $\angle B = 130^\circ – 60^\circ = 70^\circ$.

Answer: $\angle A = 60^\circ$, $\angle B = 70^\circ$, $\angle C = 50^\circ$

Ans: D

Solution:
Let $\angle A = x$, $\angle B = y$, $\angle C = z$.
We have $x – y = 30^\circ$ and $y – z = 15^\circ$.
Also, $x + y + z = 180^\circ$.
From $x – y = 30^\circ$, we have $x = y + 30^\circ$.
From $y – z = 15^\circ$, we have $z = y – 15^\circ$.
Substitute $x$ and $z$ in $x + y + z = 180^\circ$:
$(y + 30^\circ) + y + (y – 15^\circ) = 180^\circ$
$3y + 15^\circ = 180^\circ$
$3y = 165^\circ$
$y = 55^\circ$
Therefore, $\angle B = 55^\circ$.

Answer: $55^\circ$

Ans: D

Solution:
Since AD is the angle bisector of angle A and AD is perpendicular to BC, we have two congruent triangles, ABD and ACD (ASA congruence: angle BAD = angle CAD, AD=AD, angle ADB = angle ADC = 90 degrees). Therefore, AB = AC.

Answer: Triangle ABC is isosceles.