C-16-Q030mediumsingle_mcq

The area of an isosceles triangle with equal sides $a$ and base $b$ is —

a $\dfrac{b}{2}\sqrt{4a^2 - b^2}$
b $\dfrac{b}{4}\sqrt{4a^2 - b^2}$
c $\dfrac{a}{4}\sqrt{4a^2 - b^2}$
d $\dfrac{b}{4}\sqrt{a^2 - b^2}$

ব্যাখ্যা

The height drawn to the base bisects it, so $h=\sqrt{a^2-(b/2)^2}=\tfrac12\sqrt{4a^2-b^2}$ . Area $=\tfrac12\cdot b\cdot h=\dfrac{b}{4}\sqrt{4a^2-b^2}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-16-Q031An isosceles triangle has base cm and area sq cm. The length of each equal side is —
C-16-Q032The equal sides of an isosceles triangle are m and the area is sq m. The base is —
C-16-Q033An isosceles triangle has perimeter m, with each equal side of the base. The base is —
C-16-Q034With perimeter m and each equal side of the base, each equal side measures —
C-16-Q035Two roads make an angle at a place. Two persons walk for hours at km/h and km/h respectively. The direct dista
C-16-Q036Approximate direct distance between the two persons in the previous problem is —
C-16-Q037In right-angled triangle with right angle at , , . Then equals —
C-16-Q038With , , and , the length equals —
C-16-Q039In the same triangle, equals —
C-16-Q040The ratio of areas in the above figure is —

Chapter 16: Mensuration — সব প্রশ্ন দেখুন (210টি) →

The area of an isosceles triangle with equal sides $a$ and base $b$ is —

\dfrac{b}{2}\sqrt{4a^2 - b^2}

\dfrac{b}{4}\sqrt{4a^2 - b^2}

\dfrac{a}{4}\sqrt{4a^2 - b^2}

\dfrac{b}{4}\sqrt{a^2 - b^2}

ব্যাখ্যা

The height drawn to the base bisects it, so $h=\sqrt{a^2-(b/2)^2}=\tfrac12\sqrt{4a^2-b^2}$ . Area $=\tfrac12\cdot b\cdot h=\dfrac{b}{4}\sqrt{4a^2-b^2}$ .

The area of an isosceles triangle with equal sides aaa and base bbb is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The area of an isosceles triangle with equal sides aaa and base bbb is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The area of an isosceles triangle with equal sides $a$ and base $b$ is —

The area of an isosceles triangle with equal sides $a$ and base $b$ is —