The area of a triangle with sides \(13\) cm, \(14\) cm and \(15\) cm is
- (a)\(105\) cm²
- (b)\(91\) cm²
- (c)\(42\) cm²
- (d)\(84\) cm²
Revision notes, 17 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.
Mensuration unit: 14 of 80 theory marks (Area and Perimeter, Surface Area and Volume).
In the NCERT book Ganita Manjari: Part I, Chapter 6, Measuring Space: Perimeter and Area.
Triangle \(\dfrac12 \times \text{base} \times \text{height}\); rectangle \(lb\); parallelogram \(bh\); trapezium \(\dfrac12(a + b)h\); rhombus \(\dfrac12 d_1d_2\); equilateral triangle of side \(a\): \(\dfrac{\sqrt3}{4}a^2\). Circle \(\pi r^2\); sector of angle \(\theta^\circ\): \(\dfrac{\theta}{360}\pi r^2\).
For a triangle with sides \(a, b, c\) and semi-perimeter \(s = \dfrac{a + b + c}{2}\):
\[\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}\]
It needs only the three sides, so it works when no height is known. A quadrilateral with a known diagonal splits into two triangles.
The incircle touches all three sides; its centre is where the angle bisectors meet. Joining the incentre to the vertices splits the triangle into three triangles of height \(r\), so \(\text{Area} = rs\) and \(r = \dfrac{\text{Area}}{s}\).
For a cyclic quadrilateral with sides \(a, b, c, d\) and \(s = \dfrac{a + b + c + d}{2}\): \(\text{Area} = \sqrt{(s - a)(s - b)(s - c)(s - d)}\). Heron's formula is the case \(d = 0\).
Find the area of a triangle with sides \(13\), \(14\) and \(15\) cm.
\(s = 21\); area \(= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84\) cm².
Find the length of an arc of angle \(90^\circ\) in a circle of radius \(14\) cm (take \(\pi = \dfrac{22}{7}\)).
\(l = \dfrac14 \times 2 \times \dfrac{22}{7} \times 14 = 22\) cm.
Find the inradius of the triangle with sides \(6\), \(8\) and \(10\) cm.
Area \(= 24\) cm², \(s = 12\), so \(r = 2\) cm.
Topics in this chapter: Heron's formula · Perimeter and arc length · Areas of standard shapes · Brahmagupta's formula · Incircle.
Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.
You can use Heron's formula, circle and arc formulas and the standard area formulas in one step.
You can find missing sides from a perimeter, an inradius from Area = rs, and areas of rhombuses and isosceles triangles.
You can find the area of a quadrilateral in two triangles and answer multi-part questions on tracks, sectors and signboards.
You can use Brahmagupta's formula, altitudes from areas, and composite shapes with circles.
Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 3 of the 17 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.
The area of a triangle with sides \(13\) cm, \(14\) cm and \(15\) cm is
The perimeter of a semicircular protractor of radius \(7\) cm is (take \(\pi = \dfrac{22}{7}\))
The area of an equilateral triangle of side \(8\) cm is
Stretch yourself: original geometry problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I99 · Problem I17 · Problem I85 · Problem I92. All 28 →
Want it against the clock? Take a timed 30-mark chapter test on Area and Perimeter, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Essentials or the free trial).