Geometry – Page 72

Question-192325

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Question Number 192325 by cherokeesay last updated on 14/May/23 Answered by a.lgnaoui last updated on 14/May/23 $s urface S = \frac{1}{2} (AB \times BF) = \frac{1}{2} CD \times (BD - DF)$ $S = 2 CD$ $C alcul CD$ $△ ACE ∡ EAC = ∡ DEF = θ$ $$\boldsymbol{\mathrm{C}}\mathrm{E}=\mathrm{AEsin}\:\theta\:\:\:\mathrm{cos}\:\theta=\frac{\mathrm{5}}{\mathrm{AE}}\Rightarrow\mathrm{AE}=\frac{\mathrm{5}}{\mathrm{cos}\:\theta}…

Question-61235

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Question Number 61235 by ajfour last updated on 30/May/19 Commented by ajfour last updated on 30/May/19 $F i n d h a n d R o f m a x i m u m v o l u m e$ $i n s c r i b e d c y l i n d e r w i t h i t s b a s e o n$ $f a c e A B C o f a g e n e r a l p y r a m i d .$ Commented by…

Question-192288

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Question Number 192288 by universe last updated on 14/May/23 Answered by mahdipoor last updated on 14/May/23 $g e t ∠ B C G = a, B C = b$ $G D = G B + B D = B C . t a n a + B C . t a n (90 - a)$ $= b (t a n a + c o t a)$ $${FD}={GD}.{tana}={b}\left({tan}^{\mathrm{2}} {a}+\mathrm{1}\right)={b}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {a}}\right)…

Question-61186

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Question Number 61186 by Tawa1 last updated on 30/May/19 Commented by Prithwish sen last updated on 30/May/19 $Area of the shadded portion = Area of the incircle$ $- Area of the quarter circle + \frac{3}{4} (Area of the$ $square - Area of the incircle)$ $$=\pi\mathrm{5}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{10}\right)^{\mathrm{2}}…

Question-61169

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Question Number 61169 by Tawa1 last updated on 29/May/19 Commented by Tawa1 last updated on 29/May/19 $Find the area of the Black$ Answered by mr W last updated…

Question-192202

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Question Number 192202 by universe last updated on 11/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-192177

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Question Number 192177 by normans last updated on 10/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

cube-ABCD-EFGH-side-4-cm-point-P-is-center-BF-and-Q-center-AD-distance-E-to-PQG-

Tinku Tara June 4, 2023 Geometry 0 Comments

Question Number 126609 by abdullahquwatan last updated on 22/Dec/20 $cube ABCD . EFGH s i d e 4 c m p o i n t P i s c e n t e r B F a n d Q c e n t e r A D . d i s t a n c e E t o P Q G$ Answered by liberty last updated on 22/Dec/20 $t h e e q u a t i o n o f p l a n e p a s s e s t r o u g h t$ $P (4, 4, 2), Q (2, 0, 0), G (0, 4, 4)$ $$\Rightarrow\: $| \begin{matrix} x - 2 y z \\ - 2 4 4 \\ 2 4 2 \end{matrix} |$ =\:\mathrm{0} \…

Question-126557

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Question Number 126557 by I want to learn more last updated on 21/Dec/20 Commented by I want to learn more last updated on 21/Dec/20 $$\mathrm{Find}\:\:\mathrm{k}…

Question-60987

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Question Number 60987 by ajfour last updated on 28/May/19 Commented by ajfour last updated on 28/May/19 $F i n d t h e i l l u m i n a t e d a r e a o f$ $t h e i n n e r c u r v e d s u r f a c e o f$ $s h o w n, h o l l o w o p e n c y l i n d e r .$ $[w h e r e a > 2 R (\frac{H}{h} - 1)]$ …

Category: Geometry