Question
In ∆ABC,∠ACB is obtuse angle, seg AD ⊥ seg BC. Prove that:
AB² = BC² + AC² - 2BC×CD
AB² = BC² + AC² - 2BC×CD
✓
Answer
Get the step-by-step solution for this question inside the Vidyadip app.
Get the answer in the appNeed a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find $\frac{A(\triangle A B C)}{A(\triangle A P Q)}$
Solution :
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
[The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ \times ⬜ }{ ⬜ \times ⬜ }$
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ }{⬜}$
→Solution :
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
[The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ \times ⬜ }{ ⬜ \times ⬜ }$
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ }{⬜}$
In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar.
In ΔAPB and ΔAQC
In ΔAPB and ΔAQC
∠APB = [ ]° ......(i)
∠AQC = [ ]° ......(ii)
∠APB ≅ ∠AQC .....[From (i) and (ii)]
∠PAB ≅ ∠QAC .....[______]
ΔAPB ~ ΔAQC .....[______]
Write the correct number in the given boxes from the following A. P.
1, 8, 15, 22, . . .
Here
a = ⬜, $t _1$ = ⬜, $t _2$ = ⬜, $t _3$ = ⬜
$t _2- t _1$ = ⬜ - ⬜ = ⬜
$t _3- t _2$ = ⬜ - ⬜ = ⬜
∴ d = ⬜
→1, 8, 15, 22, . . .
Here
a = ⬜, $t _1$ = ⬜, $t _2$ = ⬜, $t _3$ = ⬜
$t _2- t _1$ = ⬜ - ⬜ = ⬜
$t _3- t _2$ = ⬜ - ⬜ = ⬜
∴ d = ⬜
Write the values of the following trigonometric ratios.
$\cos 45^{\circ}=\frac{⬜}{⬜}$
→$\cos 45^{\circ}=\frac{⬜}{⬜}$
prove the theorem Opposite angles of a cyclic quadrilateral are supplementry.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.
$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
→$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
Complete the following table using given information.
Complete the following activity to find the number of natural numbers from 1 to 171 which are divisible by 5 .
Complete the following activity to solve the simultaneous equations
Complete the following activity to find the 27 th term or the A.P. 9. 4. -1,-6,-11, ........
