## Question

secondary 3 | A Maths

Anyone can contribute an answer, even non-tutors.

##### MM

chevron_right chevron_right

Date Posted: 8 months ago
Views: 47
J
8 months ago
Use formula

Area of △ = ½ab sin c

½ (3 + √2)(b + 3√2) sin45° = 11/2 + 3√2

½ (3b + 9√2 + b√2 + 3(2) ) (√2 / 2) = 11/2 + 3√2

3b√2 / 4 + 9/2 + b/2 + 3√2 /2 = 11/2 + 3√2

9/2 + b/2 + (3b/4 + 3/2 ) √2 = 11/2 + 3√2

Comparing coefficients,

9/2 + b/2 = 11/2
b/2 = 11/2 - 9/2 = 1
b = 2

Or

Comparing coefficients,

3b/4 + 3/2 = 3

3b/4 = 3 - 3/2 = 3/2

b = 3/2 x 4/3 = 2
MM
8 months ago
I tried it but it did not work
Snell
8 months ago
0.5 x (3+√2) x (b + 3√2) x sin 45deg
= 0.5 x (3b + 9√2 + b√2 + 6) x 1/√2
= 0.5 x [3b+6 + (9+b)√2] x 1/√2
= 0.5 x [3b+b√2+ 6 +9√2] x 1/√2
= 0.5 x [3b/√2+b+ 6/√2 +9]
or
= 0.5 x [b + 9 + 6/√2 + 3b/√2]

(b + 9)/2 = 11/2
b = 2 (shown)

.
.
.
J
8 months ago
Refer to my edited working above. If there's any doubts feel free to ask.
Eric Nicholas K
8 months ago
Will write this up later on paper
MM
8 months ago
I am understood