Giúp e vs bài 13,14 mai e nộp rồi huhu

nguyen153246

2 năm trước

Loga Sinh Học lớp 12

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ankid1412

Đáp án:

Giải thích các bước giải:

gọi bán kính hình tròn là a

a, diện tích là

a x a x 3,14

mà a giảm 20 %

nên diện tích là

( a x 20 % ) x ( a x 20 % ) x 3,14

= 20 % a x 20 % a x 3,14

= 4 % x a x a x 3,14

mà diện tích là a x a x 3,14

nên bán kính giảm 20 % thì diện tích giảm 4 %

b, chu vi là a x 2 x 3,14

mà a giảm 20 %

nên chu vi là

20 % x a x 2 x 3,14

mà chu vi là a x 2 x 3,14

nên khi bán kính giảm 20 % thì chu vi giảm 20 %

Đ/S a, giảm 4 % b, giảm 20 %

bài 14 mik ko làm đc

Vote (0) Phản hồi (0) 2 năm trước

phungthimyloc2003

Bài 13:

+) Gọi bán kính hình tròn là a

a) Diện tích là

a x a x 3,14

Mà a giảm 20 %

=> diện tích là (a x 20%) x (a x 20%) x 3,14

= 20% a x 20% a x 3,14

= 4% x a x a x 3,14

Mà diện tích là a x a x 3,14

=> bán kính giảm 20% thì diện tích giảm 4%

b) Chu vi là:

a x 2 x 3,14

Mà a giảm 20%

=> Chu vi là :

20% x a x 2 x 3,14

Mà chu vi là a x 2 x 3,14

=> Khi bán kính giảm 20% thì chu vi giảm 20%

Đáp số:

a) giảm 4%

b) giảm 20%

Bài 14:

+) Xét tam giác ABC và tam giác ACD có: $\frac{A B}{C D} = \frac{4}{7}$

Khoảng cách từ A xuống DC bằng khoảng cách từ C xuống AB => $\frac{S_{A B C}}{S_{A C D}} = \frac{4}{7}$

+) Xét tam giác ABC và tam giác ACD có chung đáy AC, $\frac{S_{A B C}}{S_{A C D}} = \frac{4}{7}$

=> Khoảng cách từ B đến AC bằng $\frac{4}{7}$ khoảng cách từ D đến AC

+) Xét tam giác BMC và tam giác DMC có chung đáy MC

Khoảng cách từ B đến AC bằng $\frac{4}{7}$ khoảng cách từ D đến AC

$\frac{S_{B M C}}{S_{C M D}} = \frac{4}{7}$

=> Diện tích tam giác CMD là:

$15 : 4 \times 7 = 26, 25$ (c $c m^{2}$ )

Diện tích tam giác BCD là:

24 $15 + 26, 25 = 41, 25$ (c $c m^{2}$ )

Diện tích tam giác ABC là:

66 : 7 x 4 = 264/7 = 37,71 (c $c m^{2}$ )

Diện tích hình thang ABCD là:

37,71 + 66 = 103,71 (cm2)

Đáp số: 103,71 cm2

Vote (0) Phản hồi (0) 2 năm trước

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