Question

B

Figure 7.50
D
M
C
Figure 7.50 represents a pyramid ABCD with
a horizontal base ABC.
AB=AC = 5 cm and BD = CD = 13 cm.
D is vertically above A and ZBAD = ZCAD = 90°.
M is the midpoint of BC.
Calculate
lil the length AM
lil the angle BCD
(iii) the angle between the planes BCA and BCD.


Question 2 please

Answer 1

This is not in my area but I will try to do my best to help you, I hope so, but before responding, contact a more qualified person in this area.

Explanation:

Sure, I can help you with that. Let's start by solving each of the problems.

(i) To find the length of AM, we can use the Pythagorean theorem. Since M is the midpoint of BC, the length of BM and MC are equal. Therefore, BM = MC = 5/2 = 2.5 cm.

Now, we can use the Pythagorean theorem to find AM:

AM^2 = AB^2 - BM^2

AM^2 = 5^2 - (2.5)^2

AM^2 = 25 - 6.25

AM^2 = 18.75

AM = √18.75

AM ≈ 4.33 cm

(ii) To find angle BCD, we can use the law of cosines. Given that BD = CD = 13 cm and BC = 5 cm, we can use the law of cosines to find angle BCD:

cos(BCD) = (BD^2 + CD^2 - BC^2) / (2 * BD * CD)

cos(BCD) = (13^2 + 13^2 - 5^2) / (2 *