Find the area of the triangle whose sides are a=13 – Class 11 Mathematics | ConceptFirst

Exercise 10(a) · Q12

a=13

Telangana TsbieTextbookSubjectiveImportance★★★★★

53% · 21/40 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Compute the semi-perimeter:

s=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21.

Step 2. Compute $s-a,s-b,s-c$ :

s-a=21-13=8,\quad s-b=21-14=7,\quad s-c=21-15=6.

Step 3. Apply Heron's Formula:

$\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21\times8\times7\times6}.$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2018Set 1A7 marks
$P_1, P_2, P_3$
(i) $\dfrac{1}{P_1} + \dfrac{1}{P_2} + \dfrac{1}{P_3} = \dfrac{1}{r}$
(ii) $P_1 P_2 P_3 = \dfrac{(abc)^3}{8R^3} = \dfrac{8\Delta^3}{abc}$ .
›Reveal solutionSolution

Using Δ=½aP1=½bP2=½cP3, the altitudes are P1=2Δ/a etc.; combining with s=Δ/r and Δ=abc/4R gives both parts.

Since the area of the triangle can be computed using any side as base with its corresponding altitude:

$\Delta = \dfrac12 a P_1 = \dfrac12 b P_2 = \dfrac12 c P_3$

So: $P_1=\dfrac{2\Delta}{a},\quad P_2=\dfrac{2\Delta}{b},\quad P_3=\dfrac{2\Delta}{c}$

(i) Prove $\frac1{P_1}+\frac1{P_2}+\frac1{P_3}=\frac1r$ :

$\dfrac{1}{P_1}+\dfrac{1}{P_2}+\dfrac{1}{P_3} = \dfrac{a}{2\Delta}+\dfrac{b}{2\Delta}+\dfrac{c}{2\Delta} = \dfrac{a+b+c}{2\Delta} = \dfrac{2s}{2\Delta}=\dfrac{s}{\Delta}$

(where $s=\frac{a+b+c}2$ is the semi-perimeter). Using the standard inradius formula $r=\dfrac{\Delta}{s}$ , i.e. $\dfrac1r=\dfrac s\Delta$ :

$\dfrac{1}{P_1}+\dfrac{1}{P_2}+\dfrac{1}{P_3} = \dfrac{s}{\Delta} = \dfrac{1}{r}$

(ii) Prove $P_1P_2P_3=\dfrac{8\Delta^3}{abc}$ , and relate it to R:

$P_1P_2P_3 = \dfrac{2\Delta}{a}\cdot\dfrac{2\Delta}{b}\cdot\dfrac{2\Delta}{c} = \dfrac{8\Delta^3}{abc}$ — this part follows directly.

Now use the circumradius formula $\Delta=\dfrac{abc}{4R}$ , so $\Delta^3=\dfrac{(abc)^3}{64R^3}$ . Substituting:

…

Start your 14-day free trial to see the full solution →

←Previous question · 10.4Q11 — A man observes the angle of elevation of the top of a tower to be 45^°…→Next question · 10.4Q13 — Find the area of triangle ABC given b=10, c=12 and A=60^°.