IN KAZAKHSTAN Winter sports in Shymbulak
The Shymbulak Resort and the Mede
destinations for adventurers who are
koen en winter sports. There is show
from November until May! You can
learn to ski skate and snowboard here
ay
ADVENTURE HOLIDAYS
2
Complex
in Almaty are popular
2.
Rafting at the Sharyn river
The Sharyn river is the perfect place
for rafting. We organise day trips
to the Sharyn National Park every
weekend The trip includes a two-
hour rafting experience and a tour
of the Sharyn canyon. All our guides
speak Kazakh, Russian and English
3
Camel riding in the steppe
This is Karabura, one of our
friendly camels. She's waiting
for you to come and join her
for a ride this weekend in
the past, the nomads crossed
deserts and travelled for
months on the back of a camel
Our ride only lasts one hour,
but it's a lot of fun!
Climbing in the
Tien Shan Mountains
Do you like mountaineering? The
Tien Shan is the largest mountain
range in Central Asia and attracts
thousands of climbers every year.
Only experienced climbers can go
up the Khan Tengri, which is the
highest peak in Kazakhstan, but
trekking is suitable for everyone.​

АВСД - прямоугольник  ⇒   ∠А=∠В=∠С=∠Д=90° .

Так как МА⊥ пл. АВСД  ⇒  МА ⊥АВ , МА⊥АД , МА⊥АС.

Тогда треугольники АВМ , АДМ, АСМ, АДС, АДВ - прямоугольные , и к ним можно применить теорему Пифагора.

1)\; \; MB=\sqrt{AB^2+AM^2}=\sqrt{3^2+1^2}=\sqrt{10}2)\; \; MD=\sqrt{AD^2+AM^2}=\sqrt{4^2+1^2}=\sqrt{17}3)\; \; AC=\sqrt{AD^2+CD^2}=\sqrt{4^2+3^2}=54)\; \; BD=\sqrt{AD^2+AB^2}=\sqrt{4^2+3^2}=5\; ,\; \; AC=BD\; .

5)\; \; CM=\sqrt{AC^2+AM^2}=\sqrt{5^2+1^2}=\sqrt{26}6)\; \; S(MAC)=\frac{1}{2}\cdot AC\cdot AM=\frac{1}{2}\cdot 5\cdot 1=2,5

Пошаговое объяснение:

АВСД - прямоугольник  ⇒   ∠А=∠В=∠С=∠Д=90° .

Так как МА⊥ пл. АВСД  ⇒  МА ⊥АВ , МА⊥АД , МА⊥АС.

Тогда треугольники АВМ , АДМ, АСМ, АДС, АДВ - прямоугольные , и к ним можно применить теорему Пифагора.

1)\; \; MB=\sqrt{AB^2+AM^2}=\sqrt{3^2+1^2}=\sqrt{10}2)\; \; MD=\sqrt{AD^2+AM^2}=\sqrt{4^2+1^2}=\sqrt{17}3)\; \; AC=\sqrt{AD^2+CD^2}=\sqrt{4^2+3^2}=54)\; \; BD=\sqrt{AD^2+AB^2}=\sqrt{4^2+3^2}=5\; ,\; \; AC=BD\; .

5)\; \; CM=\sqrt{AC^2+AM^2}=\sqrt{5^2+1^2}=\sqrt{26}6)\; \; S(MAC)=\frac{1}{2}\cdot AC\cdot AM=\frac{1}{2}\cdot 5\cdot 1=2,5

Пошаговое объяснение: