In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. Find area of remaining portion of the triangle (Take √3 = 1.732).
98.55 cm²
Area of the equilateral triangle of side 24 cm is \(\frac{\sqrt3}{4}(24)^2 = 144\sqrt3 = 144\times1.732 = 249.408\ \text{cm}^2\).
The inradius of an equilateral triangle of side \(a\) is \(r=\dfrac{a}{2\sqrt3}\); for \(a=24\), \(r=\dfrac{24}{2\sqrt3}=4\sqrt3\ \text{cm}\), so \(r^2=48\).
Area of the inscribed circle is \(\pi r^2 = 48\pi\); using \(\pi=22/7\), this equals \(48\times\frac{22}{7}=\frac{1056}{7}\approx150.857\ \text{cm}^2\).
Area of the remaining (shaded) portion = area of triangle − area of circle \(= 249.408-150.857 = 98.551\approx98.55\ \text{cm}^2\).
Similar Questions
In a \(\triangle ABC\), \(\angle A:\angle B:\angle C=2:3:4\). A line (l) is drawn parallel to BA, then the \(\angle ACD\) is:
In the given \(\triangle ABC\), \(DE\parallel BC\). If \(BC=8\) cm, \(DE=6\) cm and area of \(\triangle ADE=90\) cm², then what is the area of \(\triangle ABC\) (in cm²)?
Important Questions from Triangles
Among the following options, which are NOT sides of a triangle?
In a Δ ABC, if ∠A = 120° and AB = AC, then the values of ∠B and ∠C are respectively:
In the equilateral Δ ABC, the base BC is trisected at D and E. The line through D, Parallel to AB, meets AC at F and the line through E parallel to AC meets AB at G. If EG and DF intersect at H, then what is the ratio of the sum of the area of parallelogram AGHF and the area of the Δ DHE to the area of the Δ ABC?
The product of the perimeter of a triangle, the radius of its in‐circle, and a number gives the area of the triangle. The number is
A man goes 24 m towards east and then 10 m towards north. How far is he away from his initial position?
